diff --git "a/c6h_checkpoint_8_32800.ndjson" "b/c6h_checkpoint_8_32800.ndjson" new file mode 100644--- /dev/null +++ "b/c6h_checkpoint_8_32800.ndjson" @@ -0,0 +1,18831 @@ +{ + "Tag": [ + "superior algebra", + "superior algebra unsolved" + ], + "Problem": "For which groups $\\mathcal{G}$ is $\\textrm{End}\\ \\mathcal{G},$ the set of endomorphisms of $\\mathcal{G},$ a group under composition? :)", + "Solution_1": "Only if $G$ is trivial; indeed $End(G)$ must have the identity mapping as a unit element (by unicity), and if $G \\neq \\{e\\}$, then $f(x)=e$ is not injective, i.e not bijective, and has no inverse." +} +{ + "Tag": [ + "geometry", + "3D geometry", + "probability", + "inequalities", + "triangle inequality" + ], + "Problem": "Two sides of a triangle measure 2'' and 4'' and the third side measures some whole number of inches. If a cube with faces numbered 1 through 6 is rolled, what is the probability, expressed as a common fraction, that the number showing on top could be the number of inches in the length of the third side of the triangle?", + "Solution_1": "the two shorter sides must equal the larger side or be greater than it.\r\n$ \\ 2\\plus{}1<4$\r\n$ \\ 2\\plus{}2\\equal{}4$\r\n$ \\ 2\\plus{}3>4$\r\n$ \\ 2\\plus{}4>4$\r\n$ \\ 2\\plus{}4>5$\r\n$ \\ 2\\plus{}4\\equal{}6$\r\nSo for 5 possabilities this works.\r\nAnswer is $ \\boxed {\\frac {5}{6}}$", + "Solution_2": "I believe that your solution is incorrect. Call the length of the last side $ n$. By the Triangle Inequality, $ 4 \\minus{} 2 \\equal{} 2 < n < 6 \\equal{} 4 \\plus{} 2$. Therefore, $ n$ must be either $ 3, 4$ or $ 5$. A total of $ 3$ choices, so our answer is $ \\frac {3}{6} \\equal{} \\boxed{\\frac {1}{2}}$. \r\n\r\n[hide=\"NOTE\"]\nThe sum of the lengths of two shorter sides must be numerically greater than the length of largest side for the figure to be a triangle. If the sum of the lengths of the two shorter sides is numerically [i]equals[/i] the length of the longer side, the triangle is degenerate, or simply put, a straight line.[/hide]" +} +{ + "Tag": [ + "trigonometry" + ], + "Problem": "A loaded freight car of mass 75 tonne is pulled up a 0.75% grade 1200m long by a steady drawbar pull of 4500N.The train resistance is 35N per tonne and the initial speed is 12m/s.Find the speed at the top of the incline.", + "Solution_1": "Draw the free-body diagram. If I'm understanding your problem correctly, we have three forces that we need to take into account: the pull force (up), the resistance (down), and the weight (down).\r\n\r\nTherefore, we should have $ \\sum F\\equal{}F_{p}\\minus{}F_{r}\\minus{}mg\\sin{\\theta}\\equal{}ma$ where $ \\theta$ is determined from the length of the ramp and the grade. Then you can solve for $ a$ and use kinematic equations.\r\n\r\nYou said .75% grade. That is a very small number. Are you sure you don't mean 75%?", + "Solution_2": "[quote=\"JRav\"]Draw the free-body diagram. If I'm understanding your problem correctly, we have three forces that we need to take into account: the pull force (up), the resistance (down), and the weight (down).\n\nTherefore, we should have $ \\sum F \\equal{} F_{p} \\minus{} F_{r} \\minus{} mg\\sin{\\theta} \\equal{} ma$ where $ \\theta$ is determined from the length of the ramp and the grade. Then you can solve for $ a$ and use kinematic equations.\n\nYou said .75% grade. That is a very small number. Are you sure you don't mean 75%?[/quote]\r\nI'm sure about this. And your solution is completely right! :wink: . But I'm curious whether we can apply the theorem:\"Conservation of energy\" in this case or not?I applied it and got the double result compared to the true result(not sure it will work since we have the force of the train resistance,which I think is a kind of friction).", + "Solution_3": "I haven't worked out any of the numbers so you'll have to tell me what it should be. Show me your work for conservation of energy and I'll try to spot your error.", + "Solution_4": "[hide]Here's my solution:\nApplying the conservation energy into two moments: 1. car of mass is pulled from rest\n 2.At the top of the incline\nWe have: 1/2.m.v_1^2+4500.1200=75.35+1/2.m.v_2^2+mgh (choosing the origin as the place where car is pulled from rest)\nNow plugging in v_1=12,m=75000,g=9,8 and h=1200.(0,75%)=9 into this equation above, we will get: v_2=10.6(rounded result)\n\nYour solution:\nWe will easily show that: a=-97/2000\nThen applying the formula: v^2=v_(0)^2+2as we will get: v=5.3m/s(rounded result). It's also the right result![/hide]", + "Solution_5": "I can't quite understand all of your steps because I can't tell what all the numbers are. Just clarify your numbers for me.", + "Solution_6": "[quote=\"ghjk\"][hide]Here's my solution:\nApplying the conservation energy into two moments: 1. car of mass is pulled from rest\n 2.At the top of the incline\nWe have: 1/2.m.v_1^2+4500.1200=75.35+1/2.m.v_2^2+mgh (choosing the origin as the place where car is pulled from rest)\nNow plugging in v_1=12,m=75000,g=9,8 and h=1200.(0,75%)=9 into this equation above, we will get: v_2=10.6(rounded result)\n\nYour solution:\nWe will easily show that: a=-97/2000\nThen applying the formula: v^2=v_(0)^2+2as we will get: v=5.3m/s(rounded result). It's also the right result![/hide][/quote]\r\nIn my equation, m is the weight of a loaded freight of car.\r\n v_1 is the initial speed\r\n 4500.1200 is the work of the drawbar pull\r\n h is the height when the car is at the top of the inclined plane\r\n 75.35.[b]1200[/b](I forget to multiply by 1200) is the work of the train resistance" +} +{ + "Tag": [ + "function", + "algebra unsolved", + "algebra" + ], + "Problem": "Let a1,a2,\u30fb\u30fban be real numbers and these are different each other. This means \"if ai=aj,then i=j\".\r\nIt is possivle to describe k-th largest number(Xk) among a1,a2,\u30fb\u30fb,an , only by using four arithmetic operations (+-*/) and absolute value(abs)?\r\nIt is permitted to use \"max(b1,b2,\u30fb\u30fbbj) \" and \"min(b1,b2\u30fb\u30fb,bj)\", because these can be written by +-*/ and abs.\r\nFor example ,max(x,y)=(x+y+abs(x-y))/2 , min(x+y)=(x+y-abs(x-y))/2 , max(x,y,z)=max(max(x,y),z) , min(x,y,z)=min(min(x,y),z) and so on by induction.\r\n\r\nWhen n=2. X1=max(a1,a2) and X2=min(a1,a2).\r\nWhen n=3, X1=max(a1,a2,a3), X3=min(a1,a2,a3) and X2=a1+a2+a3-max(a1,a2,a3)-min(a1,a2,a3).\r\nWhen n=4 X1=max(a1,a2,a3,a4) X4=min(a1,a2,a3,a4).\r\n About X2 and X3, solve the simultaneous equation X2+X3=a1+a2+a3+a4-max(a1,a2,a3,a4)-min(a1,a2,a3,a4) and X2^2+X3^2=a1^2+a2^2+a3^2+a4^2-max(a1,a2,a3,a4)^2-min(a1,a2,a3,a4)^2 and let the large root X2,smaller root X3.\r\nIt there better solutions for n=4?\r\nAnd what about n>4?", + "Solution_1": "what problem is that?!\r\nif you admit that you can use \"max\" function then you can order all!! :huh:", + "Solution_2": "The max function does not (directly) tell you, for example, the second or third largest number." +} +{ + "Tag": [], + "Problem": "$ a^2\\equal{}n^6\\plus{}1999$\r\nsolve in integers", + "Solution_1": "Nope ,I dont think there are any integer solutions .\r\nwe have \r\n$ (a - n^{3})(a + n^{3}) = 1999$\r\nSince $ 1999$ is a prime number .\r\n$ (a + n^{3}) = 1999$ and $ (a - n^{3} = 1$ \r\nUsing the two equations we \r\nget $ a = 1000$ and $ n = 999^{\\frac{1}{3}}$ \r\nwhich is not an integer ." +} +{ + "Tag": [ + "LaTeX", + "inequalities unsolved", + "inequalities" + ], + "Problem": "For any positive integer $n>2$, and a sequence of positive reals $a_{1}, a_{2}, a_{3}, ....., a_{n}$; such that:\r\n$(a_{1}^{2}+.......+a_{n}^{2})^{2}> (n-1)(a_{1}^{4}+.....+a_{n}^{4})$ \r\n\r\nprove that for $1\\leq i (n-1)\\sum_{i=1}^{n}a_{i}^{2}$\r\nso for some fixed $i$ this gives: $\\ a_{i}^{2}+\\ a_{i}(\\sum_{j=1j\\not= i}^{n}a_{j}) > (n-1)\\ a_{i}^{2}$, or \r\n$\\ a_{i}+\\sum_{j=1j\\not= i}^{n}a_{j}> (n-1)\\ a_{i}$\r\nso we have $\\boxed{\\sum_{j=1}^{n}a_{j}> (n-1)a_{i}}$ for any $\\ i$ wich gives $\\ n$ sets of cyclics inequalities.\r\n\r\nNow let $\\ a_{n}\\geq a_{n-1}\\geq ...\\geq a_{1}$\r\nwe see that $\\ (k-1)a_{k}\\geq a_{1}+a_{2}+...a_{k-1}$ (1)\r\nSo for some $\\ n\\geq k>j>i$ we have: $\\sum_{l=1}^{n}a_{l}> (n-1)a_{k}$ and using (1)\r\n$\\ (k-2)a_{k}+a_{j}+a_{i}+(a_{k+1}+a_{k+2}+...+a_{n})>(n-1)a_{k}$ (2)\r\n\r\nWith $\\sum_{l=1}^{n}a_{l}> (n-1)a_{n}$\r\nthis gives using (1) again: $\\ (n-2)a_{n}+a_{j}+a_{i}>(n-1)a_{n}$\r\n\r\nor $\\ a_{j}+a_{i}>a_{n}$ (3)\r\n\r\n$\\ (n-k)(a_{j}+a_{i})>(n-k)a_{n}>a_{k+1}+a_{k+2}+..+a_{n}$ (4)\r\n\r\nCombining (2), (3) and (4):\r\n\r\n$\\ (n-k)(a_{j}+a_{i})+a_{j}+a_{i}>(n-k+1)a_{k}$\r\n\r\n$\\ (n-k+1)(a_{j}+a_{i})>(n-k+1)a_{k}$\r\n\r\nFinally $\\ a_{j}+a_{i}>a_{k}$ \r\nbecause the equations are cyclic we have:\r\n$\\ a_{j}+a_{k}>a_{i}$\r\n$\\ a_{i}+a_{k}>a_{j}$\r\nNow make $\\ a_{m}\\rightarrow a_{m}^{2}$\r\n\r\n$\\ a_{j}^{2}+a_{i}^{2}>a_{k}^{2}$\r\n$\\ a_{j}^{2}+a_{k}^{2}>a_{i}^{2}$\r\n$\\ a_{i}^{2}+a_{k}^{2}>a_{j}^{2}$\r\n\r\nand those equations are still the conditions for $\\ (a_{i},a_{j},a_{k})$ be the lengths of the sides of a triangle.", + "Solution_2": "Euh can somebody tell me if my solution is good? :)", + "Solution_3": "i think is correct" +} +{ + "Tag": [ + "LaTeX", + "topology", + "factorial", + "trigonometry", + "algebra", + "function", + "domain" + ], + "Problem": "Welcome to 2+2=5 The game where it's right to be wrong. The whole object is to answer the question above you wrong and create an easy question for the next player.\r\nFirst question is b a letter?", + "Solution_1": "nope.\r\n\r\nIf this statement is true, what is this statement?", + "Solution_2": "New York City\r\n\r\nIf three trendifs make a minfat, what is seven plus seven?", + "Solution_3": "Answer: r15s11z55y89w21\r\n\r\nWhat day is it?", + "Solution_4": "Yesterday.\r\n\r\nWhat did you have for dinner yesterday?", + "Solution_5": "Math Textbooks.\r\n\r\nWhat is your favorite game?", + "Solution_6": "Dying.\r\n\r\nWho am I?", + "Solution_7": "A math textbook", + "Solution_8": "no\r\n\r\nAm I a human?", + "Solution_9": "no you are a car\r\n2*2=", + "Solution_10": "426\r\n\r\nWhy do I smell like strawberries?", + "Solution_11": "17\r\n2*82108938239992839744683628=", + "Solution_12": "A tarantula. A big hairy tarantula...\r\n\r\n\r\nWhat is the meaning of life?", + "Solution_13": "Life has no meaning\r\nwhat is the alphabet?", + "Solution_14": "$ \\infty$\r\n\r\nDefine a piano.", + "Solution_15": "You don't have one.\n\n\nWhy am I here?", + "Solution_16": "because it's Monday.\n\nhow old am i?", + "Solution_17": "[hide=\"Don't click\"]A negative century[/hide]\n\nCan you fly?", + "Solution_18": "I'm a snail.\n\nDo I like salt?", + "Solution_19": "There is no movie.\n\nWho is in my avatar?", + "Solution_20": "jackie chan.\n\nwhat is 1^6?", + "Solution_21": "Over 9000!!!!!!!!\n\nVegeta! What does the scouter say about his power level?", + "Solution_22": "It's about 9000.\n\nVegeta! What is the scouter's name?", + "Solution_23": "Shirley Shirley Bo-Birley Banana Fana Fo Firley Me Mi Mo Mirley - Shirley!\n\nWho is N?", + "Solution_24": "nobody\n\nWho is nobody?", + "Solution_25": "Odysseus\n\nGive an example of a string literal.", + "Solution_26": "=7!\n\nIs the world a world?", + "Solution_27": "lol\n\n\n\nAm I lying?", + "Solution_28": "cheesburgers\nIs 1+5134624365 hamburgers?", + "Solution_29": "MODS PLZ LOCK, TOO MUCH SPAMMING\nmusicnmath" +} +{ + "Tag": [ + "function", + "inequalities", + "algebra", + "polynomial", + "absolute value" + ], + "Problem": "Hello everyone,\r\n\r\nI arrived at wrong answers in my attempt to solve this problem and would appreciate help.\r\n\r\nThank you very much!\r\n\r\n---\r\n\r\n1. Objects A and B move along a straight line. Their positions, [i]s[/i], with respect to an origin, at [i]t[/i] seconds, are modelled by the following functions:\r\n\r\nObject A: $ s(t) \\equal{} \\frac {7t}{t^2 \\plus{} 1}$\r\n\r\nObject B: $ s(t) \\equal{} t \\plus{} \\frac {5}{t \\plus{} 2}$\r\n\r\nWhen is object A closer to the origin than object B?\r\n\r\n---\r\n\r\nI interpreted object A's higher proximity to the origin as:\r\n\r\n$ \\left | \\frac {7t}{t^2 \\plus{} 1} \\right | < t \\plus{} \\frac {5}{t \\plus{} 2}$\r\n\r\nTo solve this inequality, I multiplied both sides by $ (t \\plus{} 2)^2 \\times (t^2 \\plus{} 1)^2$ and then isolated the terms:\r\n\r\n$ (t \\plus{} 2)^2(7t)(t^2 \\plus{} 1) \\minus{} (t^2 \\plus{} 2t \\plus{} 5)(t \\plus{} 2)(t^2 \\plus{} 1)^2 < 0$\r\n\r\n$ (t \\plus{} 2)(t^2 \\plus{} 1)[(t \\plus{} 2)7t \\minus{} (t^2 \\plus{} 1)(t^2 \\plus{} 2t \\plus{} 5)] < 0$\r\n\r\n$ (t \\plus{} 2)(t^2 \\plus{} 1)[ \\minus{} t^4 \\minus{} 2t^3 \\plus{} t^2 \\minus{} 16t \\minus{} 5] < 0$\r\n\r\n$ t \\equal{} \\minus{} 2, \\minus{} \\sqrt {1}, \\sqrt {1}, \\minus{} 0.304, \\minus{} 3.48$ \r\n\r\nI used my graphing calculator to find the regions of the graph below 0: $ x \\epsilon ( \\minus{} \\infty, \\minus{} 3.48) \\cup ( \\minus{} 0.304, \\infty)$. However, they are not the correct answers.", + "Solution_1": "Multiplying by $ t \\plus{} 2$ changes the sign of the inequality if $ t \\plus{} 2$ is negative. This shouldn't matter if $ t$ is positive, but it does mean that you need to ignore the part of your answer that occurs when $ t$ is negative (since it doesn't make physical sense). But unless you show more of your work, it's hard to tell where you could've gone wrong.", + "Solution_2": "Thanks for your reply, t0rajir0u.\r\n\r\nI have edited my post to show more steps. Please let me know if more steps are required.", + "Solution_3": "Notes:\r\n\r\n1) There was never any need to multiply by $ (t^2 \\plus{} 1)^2$. I presume the squaring was to avoid multiplying by a negative, but $ t^2 \\plus{} 1 > 0$ for all real $ t$.\r\n\r\n2) More generally, you're searching for the points where the LHS and RHS are equal. Start off with this explicit goal (\"First, we find the points where the two particles are equidistant from the origin.\") and then solve the resulting [i]equation[/i]. After this, you divide everything into intervals and can determine which function is closer to 0 on each interval.\r\n\r\n3) Your behavior with absolute value signs is very odd -- some of them (but not as many as I would expect) are there, but then they mysteriously disappear. Either decide that $ t \\geq 0$ (suggested by the fact that $ t$ is refered to as \"time\" in the problem, as t0r mentioned) or be more careful with this (e.g., start by comparing the squares of the two terms, an idea that you seem to have not-quite-fully grasped). But this is not relevant if you're going to follow the advice in the previous point.\r\n\r\n4) You've botched your polynomial multiplication/addition. (Why didn't you check this more carefully? It's the obvious place for something to have gone wrong!) You also solved at least one of your equations incorrectly (the roots of $ t^2 \\plus{} 1 \\equal{} 0$ are $ \\pm \\sqrt {1}$ ?!) but at that point it didn't matter anymore.", + "Solution_4": "Thanks for your response, JBL.\r\n\r\nThe first part of this question asked to determine the times when the objects are at the same position. I did this successfully to get:\r\n\r\nEquation = $ t^4 \\plus{} 2t^3 \\minus{} t^2 \\minus{} 12t \\plus{} 5 \\equal{} 0$\r\n\r\nTherefore, $ t \\equal{} 0.417, 1.71$.\r\n\r\nThe different sections would be: $ (\\minus{}\\infty,0.417) \\cup (0.417, 1.71) \\cup (1.71, \\infty)$. However, how would I determine the regions in which object A is closer to (0,0)?", + "Solution_5": "By continuity* (the intermediate value theorem or some fairly immediate consequence of it), on any interval between meeting times the same one of the two is closer. So pick representative points and test.\r\n\r\n* Note that one of the two functions is not continuous on $ \\bf R$, so you need to add its point of discontinuity as a \"critical point,\" creating another interval." +} +{ + "Tag": [ + "algorithm", + "ratio", + "geometric series", + "geometric sequence" + ], + "Problem": "Does anyone know how to do this problem quickly?\r\n\r\n1/2^1 + 1/2^2 + 1/2^3 +...+ 1/2^9 + 1/2^10\r\n\r\n\r\nAlso is there an algorithm to add large numbers quickly?[/img][/quote]", + "Solution_1": "OK. This is known as a [i]geometric series[/i], where successive terms have a constant ratio. In this case, the ratio is $ \\frac 12$.\r\n\r\nThere are two ways to approach this:\r\n\r\n[hide=\"Geometric Sum Formula\"]\nOne can always use the geometric sum formula, which states that\n\\[ \\sum_{i \\equal{} 1}^n r^i \\equal{} \\frac {r \\minus{} r^{n \\plus{} 1}}{1 \\minus{} r}\n\\]\nIf the sigma symbol ($ \\sum$) confuses you, this is just basically\n\\[ r \\plus{} r^2 \\plus{} r^3 \\plus{} \\dots \\plus{} r^n \\equal{} \\frac {r \\minus{} r^{n \\plus{} 1}}{1 \\minus{} r}\n\\]\nSo we plug in $ r \\equal{} \\frac 12$ and $ n \\equal{} 10$, and we get $ \\frac {\\frac 12 \\minus{} \\frac 1{2048}}{\\frac 12} \\equal{} \\frac {1023}{1024}$[/hide]\n\n[hide=\"Method #2\"]\nWe could notice that if we convert all the denominators to the highest power, which is $ 2^{10}$, then we could see that the numerators are in a geometric progression:\n\\[ 512 \\plus{} 256 \\plus{} 128 \\plus{} 64 \\plus{} 32 \\plus{} 16 \\plus{} 8 \\plus{} 4 \\plus{} 2 \\plus{} 1\n\\]\nAnd since $ 1 \\plus{} 2 \\plus{} 4 \\plus{} 8 \\plus{} \\dots \\plus{} 2^k \\equal{} 2^{k \\plus{} 1} \\minus{} 1$, the numerator is 1023, and the fraction is $ \\frac {1023}{1024}$.[/hide]\r\n\r\nEither way, the answer is $ \\boxed {\\frac {1023}{1024}}$.", + "Solution_2": "can you please prove this formula?", + "Solution_3": "I think [url=http://en.wikipedia.org/wiki/Geometric_series]this[/url] might help.", + "Solution_4": "[quote=\"Poincare\"]can you please prove this formula?[/quote]\r\n\r\nSimple. Multiply the sum by $ (1 - r)$. What we get is\r\n\r\n$ \\begin{align*}(1 - r)\\sum_{i = 1}^n r^i & = (r + r^2 + \\dots + r^n) - (r^2 + r^3 + \\dots + r^{n + 1}) \\\\\r\n& = r - r^{n + 1}$\r\n\r\nObviously, we multiplied this by $ (1 - r)$, so we now divide by $ (1 - r)$ to obtain the formula.\r\n\\[ \\sum_{i = 1}^n r^i = \\frac {r - r^{n + 1}}{1 - r}\r\n\\]\r\n:D :D :D", + "Solution_5": "thx, is this in intro algebra (i wanna look it up)?" +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "If $ 0\\le x,y,z \\le 1$ Prove inequality ?\r\n\r\n$ \\frac{x}{7\\plus{}y^3\\plus{}z^3}\\plus{}\\frac{y}{7\\plus{}x^3\\plus{}z^3}\\plus{}\\frac{z}{7\\plus{}x^3\\plus{}y^3}\\le \\frac{1}{3}$", + "Solution_1": "[hide=\"Solution one\"]\n$ (x \\minus{} 1)^2(x \\plus{} 2) \\ge 0$\n\n$ \\therefore x^3 \\minus{} 3x \\plus{} 2 \\ge 0$\n\n$ \\therefore x^3 \\plus{} y^3 \\plus{} z^3 \\minus{} 3x \\minus{} 3y \\minus{} 3z \\plus{} 6 \\ge 0$\n\n$ \\therefore x^3 \\plus{} y^3 \\plus{} z^3 \\plus{} 6 \\ge 3(x \\plus{} y \\plus{} z)$\n\nSince $ 0 \\le x,y,z \\le 1 \\Longrightarrow 0 \\le x^3,y^3,z^3 \\le 1$\n\n$ \\therefore \\sum \\frac {x}{7 \\plus{} y^3 \\plus{} z^3} \\le \\sum \\frac {x}{6 \\plus{} x^3 \\plus{} y^3 \\plus{} z^3} \\le \\sum \\frac {x}{3(x \\plus{} y \\plus{} z)} \\equal{} \\frac {1}{3}$\n\nEquality when $ x \\equal{} y \\equal{} z \\equal{} 1$ [/hide]" +} +{ + "Tag": [ + "superior algebra", + "superior algebra unsolved" + ], + "Problem": "Let $R$ a ring, $S$ a multiplicatively closed subset of $R$. Prove that\r\n$S^{-1}R\\subset Q(R).$", + "Solution_1": "does $Q(R)$ mean the total quotient ring?", + "Solution_2": "Yes, that's what I meant. :wink:" +} +{ + "Tag": [], + "Problem": "If $ y\\triangleleft\\equal{}y^2\\minus{}1$, find $ (9\\triangleleft)\\triangleleft$.", + "Solution_1": "It's $ (9^2\\minus{}1)^2\\minus{}1\\equal{}6399$", + "Solution_2": "That is equivalent to 80^2-1=6400-1=6399" +} +{ + "Tag": [ + "geometry", + "perimeter" + ], + "Problem": "The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989 \\ \\text{cm}$. The length of each side of the triangle exceeds the length of each side of the square by $d \\ \\text{cm}$. The square has perimeter greater than 0. How many positive integers are NOT possible value for $d$?\r\n\r\n$\\text{(A)} \\ 0 \\qquad \\text{(B)} \\ 9 \\qquad \\text{(C)} \\ 221 \\qquad \\text{(D)} \\ 663 \\qquad \\text{(E)} \\ \\text{infinitely many}$", + "Solution_1": "[hide]Let $x$ be the length of one side of the square. The relationship between the perimeters gives $3(x+d)=4x+1989$. This gives $x=3d-1989$. Since the perimeter of the square must be greater than 0, $x$ must be greater than 0 as well. So $3d-1989>0$. This gives $d>663$, so there are $\\boxed{663}$ numbers that cannot be $d$.[/hide]", + "Solution_2": "[hide=\"Answer\"]$s_1=s_2+d$\n$3s_1=4s_2+1989=3(s_2+d)\\Rightarrow s_2=3(d-663)$\nIf $s_2>0$, then $d>663$, so we have $663$ positive integers that cannot be possible values of $d$. $\\boxed{D}$[/hide]", + "Solution_3": "[hide=\"answer\"]\nD is for Daesun (my name :P ), i'll pick D\n[/hide]" +} +{ + "Tag": [ + "probability", + "geometry", + "perimeter", + "ratio", + "rectangle", + "angle bisector", + "area of a triangle" + ], + "Problem": "[hide=\"TODAY's CONTEST\"]Today's contest\n1. A right triangle has one leg which equals 91 and hypotenuse 109. What is the length of the other leg? (1 point)\n2. What is the length of a median to side c of a triangle with sides 5, 12, and side c=8? (1 point)\n3. what is 777 to the 3rd power? (1 point)\n4. In a quadrilateral with perpendicular diagonals w= 1, x=2, y=3, and z=how much? (3 points)\n5. radius of a circle with circumference 300 pi?(1 point)\n6. What is the probability of drawing an ace or a heart in a deck of cards that is standard? (2 points)\n7. What is the area of a triangle with lenghts 7, 17, and 19? (hero's formula) (1 point)[/hide]\n[hide=\"October 22 contest Day 1\"]Day 1. Please post answers within 1 day.\n\n1. A polygon has an apothem ( a word i learned today ) of 15 and half of its perimeter is 20. What is the area of the polygon?\n\n2. What is the altitude to the hypotenuse of a right triangle with legs 3 and 4? \n\n3. What is the length of the angle bisector to Side C in a triangle where A equals 5, B equals 14 , and C equals 17? \n\n4. What is the sum of this infinite sequence. The first term is 7, and the common ratio is 3/5. \n\n5. If Josephine wants to spell out BOOTY in as many ways as possible using all letters once, how many ways can she spell it out?\n\n6. What is the angle of the shorter angle on a clock at 7:22 PM?\n\n7. How many rectangles are there in a 12 by 10 grid? 12 boxes by 10 boxes basically. [/hide]", + "Solution_1": "[hide=\"1\"]$A=\\boxed{300}$[/hide]\n\n[hide=\"2\"]$\\text{Altitude}\\,\\approx\\boxed{4.1176}$[/hide]\n\n[hide=\"3\"]$L=\\boxed{3.8\\overline{63}}$[/hide]\n\n[hide=\"4\"]$S=\\boxed{\\frac{35}{2}}$[/hide]\n\n[hide=\"5\"]$\\boxed{60\\,\\text{arrangements}}$[/hide]\n\n[hide=\"6\"]$\\text{m}\\angle=\\boxed{89^{\\circ}}$[/hide]\n\n[hide=\"7\"]$\\boxed{4290\\,\\text{rectangles}}$[/hide]", + "Solution_2": "[quote=\"i_like_pie\"][hide=1]\nNumber 5 doesn't make sense.[/quote]\r\nactually it does.\r\nit means in how many ways can you arrange the letters B,O,O,T,Y\r\nfor example, BOOTY,BOYTO,OYOTB are some arrangments", + "Solution_3": "[hide=\"4\"]\nInfinite sequences with a common ratio are geometric in the form\n$a, ar, ar^{2}, ar^{3}, ar^{n-1}$\nwhere a is the first term and r is the common ratio.\n\nThe sum of an geometric series can be represented by \n$\\frac{a}{1-r}$\n\nas long as $r \\le 1$\n\nOur sum can be represented by\n$\\frac{7}{(1-\\frac{3}{5})}$\nwhich evaluates to $\\boxed{17.5}$\n[/hide]", + "Solution_4": "[hide=\"scores and answers\"]\nFinal Scores: only 2 people DAY 1 :( \n\nI like pie-6 points\ndiscreetfouriertransform- 1 point\n\nDay 1 Winner- I_Like_Pie\n\nAnswers: \n1. 300\n2. 12/5\n3. 12/19 root 35\n4. 17.5 or 35/2\n5. 19\n6. 89 degrees\n7. 4290 rectangles\n[/hide]", + "Solution_5": "[quote=\"now a ranger\"]Today's contest\n\n1. A right triangle has one leg which equals 91 and hypotenuse 109. What is the length of the other leg? (1 point)\n2. What is the length of a median to side c of a triangle with sides 5, 12, and side c=8? (1 point)\n3. what is 777 to the 3rd power? (1 point)\n4. In a quadrilateral with perpendicular diagonals w= 1, x=2, y=3, and z=how much? (3 points)\n5. radius of a circle with circumference 300 pi?(1 point)\n6. What is the probability of drawing an ace or a heart in a deck of cards that is standard? (2 points)\n7. What is the area of a triangle with lenghts 7, 17, and 19? (hero's formula) (1 point)[/quote]\r\nDay 2:\r\n\r\n[hide=\"1\"]$60$[/hide]\n\n[hide=\"2\"]$\\frac{\\sqrt{274}}{2}$[/hide]\n\n[hide=\"3\"]$777^{3}=469097433$[/hide]\n\n[hide=\"4\"]$z=4$[/hide]\n\n[hide=\"5\"]$r=150$[/hide]\n\n[hide=\"6\"]$\\frac{4}{13}$[/hide]\n\n[hide=\"7\"]$A\\approx59.22$[/hide]", + "Solution_6": "I like pie- 10 points \r\n\r\n\r\nWinner- I like pie :) :)\r\n\r\nnooooooo moderator! ahhhhhhhhhhhh I Can't edit my first post which contains daily questions! grrrrrrrr", + "Solution_7": "Next set please?", + "Solution_8": "okay. I can't believe i can't edit my 1st post, makes it so messy grr ok next sets coming up", + "Solution_9": "1. What is the length of the common external tangent of two circles with distance 11 and radiis 5 and 15? \r\n2. What is the sum of all the integer factors of 3024?\r\n3. Twins Paul and Daniel love pies, and each draw 3 pies with radius 5. What is the perimeter of all 3 pies if none are touching? \r\n4. If there are 15 FCCLA members, and you can choose any 4, how many ways can you choose people, in any order? \r\n5. What is the square root of 39601?\r\n6. What is the area of a circumscribed quadrilateral with sides 3,4,6, and 5?\r\n7. Bob is tied by a leash to a corner inside a barn that measures 18 by 6 root 3. What area can he cover if the leash length is 12?\r\n\r\nTry to not use calculators, but it is allowed but recommended no.", + "Solution_10": "[quote=\"now a ranger\"]1. What is the length of the common external tangent of two circles with distance 11 and radiis 5 and 15? \n2. What is the sum of all the integer factors of 3024?\n3. Twins Paul and Daniel love pies, and each draw 3 pies with radius 5. What is the perimeter of all 3 pies if none are touching? \n4. If there are 15 FCCLA members, and you can choose any 4, how many ways can you choose people, in any order? \n5. What is the square root of 39601?\n6. What is the area of a circumscribed quadrilateral with sides 3,4,6, and 5?\n7. Bob is tied by a leash to a corner inside a barn that measures 18 by 6 root 3. What area can he cover if the leash length is 12?\n\nTry to not use calculators, but it is allowed but recommended no.[/quote]\r\n[hide=\"2\"]$9920$[/hide]\n[hide=\"3\"]$60\\pi$[/hide]\n[hide=\"4\"]$1365$[/hide]\n[hide=\"5\"]$\\sqrt{39601}=\\boxed{199}$[/hide]\r\n\r\nI think I won already... :D", + "Solution_11": "yeah you won\r\n\r\nwinner- I like pie\r\nHEre are my answers.\r\n\r\n1. square root of 21\r\n2. 8549\r\n3. 30 pie\r\n4. 32760 (im not sure if this is right) someone correct me if its swrong\r\n5. 199\r\n6. 6root10", + "Solution_12": "[quote=\"now a ranger\"]yeah you won\n\nwinner- I like pie\nHEre are my answers.\n\n1. square root of 21\n2. 8549\n3. 30 pie\n4. 32760 (im not sure if this is right) someone correct me if its swrong\n5. 199\n6. 6root10[/quote]\r\nNumber 3 is incorrect. The problem says they draw 3 pies [b]each,[/b] although the actual question says 3 altogether. :huh:", + "Solution_13": "oh my mistake \r\n\r\noops.. :(" +} +{ + "Tag": [ + "probability" + ], + "Problem": "If Michael rolls three fair dice, what is the probability that he will roll at least two 1's? Express your answer as a common fraction.", + "Solution_1": "1) probability that he will roll exactly 2 1's\r\n\r\n$ 3*\\frac{6}{216}\\equal{}\\frac{18}{216}$\r\n\r\n2) probability that he will roll 3 1's\r\n\r\n$ \\frac{1}{216}$\r\n\r\n$ \\frac{18}{216}\\plus{}\\frac{1}{216}\\equal{}\\frac{19}{216}$", + "Solution_2": "Wouldn't the answer be 2/27?\r\n\r\nYou can get:\r\n1 1 2\r\n1 1 3\r\n1 1 4\r\n1 1 5 \r\n1 1 6\r\n\r\nEach of which has 3 distinct arrangements. \r\n\r\nYou can also get:\r\n1 1 1\r\n\r\nWhich brings the total number of favorable outcomes to 16. 16/216 = 2/27, so the probability is 2/27.", + "Solution_3": "Hmm... Let me see...\r\n\r\nThe only way its possible is:\r\n11_\r\n_11\r\n1_1\r\n111\r\n\r\nEach of the first three has a probability of (1/6)^2*5/6=5/216.\r\n\r\nThe last one has a probability of (1/6)^3=1/216.\r\n\r\nThus the total probability is 3(5/216)+1/216=16/216=2/27, so I think you are right.", + "Solution_4": "ya\r\nthe answer is 2/27\r\ni think what fantasy did was he also included 111 as choices for the two 1's\r\nif we do it fantasy's way, we would have\r\n18+1-3=16", + "Solution_5": "oops ur right :D \r\n\r\nblah i overcounted 111 3 times..." +} +{ + "Tag": [], + "Problem": "Fie $a_{1}, a_{2},...,a_{n}$ termenii unei progresii aritmetice. Sa se arate ca pentru orice $k$ natural exista $k$ termeni dintre : $\\frac{1}{a_{1}}, ...\\frac{1}{a_{n}}$ in progresie artimetica. (nu neaparat consecutivi)", + "Solution_1": "Miros ceva putred pe aici.\r\n\r\nFie $a_{n}= k+n \\cdot r$. Avem $\\frac1{a_{n}}-\\frac1{a_{m}}= \\frac{(m-n)r}{(k+nr)(k+mr)}$.\r\n\r\nDac\u0103 $\\frac1{a_{m}}, \\, \\frac1{a_{n}}, \\, \\frac1{a_{p}}$ sunt \u00een progresie aritmetic\u0103, atunci \\[\\frac{(m-n)r}{(k+nr)(k+mr)}= \\frac{(n-p)r}{(k+pr)(k+nr)},\\] adica $\\frac{m-n}{k+mr}= \\frac{n-p}{k+pr}\\, \\, \\Longleftrightarrow \\, \\, k(m-2n+p) = r \\left( nm-2mp+pn \\right)$.\r\nDaca $k$ si $r$ sunt liniar independente peste $\\mathbb Q$, atunci $n = \\frac{m+p}2$ si $n(p+m) = 2mp$. F\u0103c\u00e2nd c\u00e2teva calcule reiese $m = p$ :maybe: Contradic\u0163ie!\r\n\r\nSper c\u0103 n-am gre\u015fit la calcule...", + "Solution_2": "Pentru progresii formate din numere intregi este adevarat, totusi." +} +{ + "Tag": [ + "logarithms", + "limit", + "algebra unsolved", + "algebra" + ], + "Problem": "Is it true that any set of positive density of natural numbers must contain infinitely many pairs $(x,y)$ of different elements such that x divides y? This started from a problem posted by Myth:if the sequence $(a_n)$ of natural numbers is increasing and $a_{n+1}-a_n$ is bounded, then this sequence has the described property. Does anyone know some results related to this problem?", + "Solution_1": "This is just to provide a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=a_n&t=2497]link to Myth's post[/url] mentioned by harazi...", + "Solution_2": "I think the answer is yes.\r\n\r\nHere is what I'm reading in :\r\nR.K. Guy, 'Unsolved problems in number theory', Springer\r\n\r\nIf no member of the sequence $\\{a_i \\}$ divides the product of $r$ other terms, Erd\u00f6s shows that \r\n$\\pi (x) + c_1 \\frac {x^{\\frac 2 {r+1}}} {(\\ln x)^2} < A(x) <\\pi (x) + c_2 \\frac {x^{\\frac 2 {r+1}}} {(\\ln x)^2} $\r\n\r\nwhere $A(x)$ is the number of terms not exceeding $x$.\r\n\r\nThus, if $S$ is a set containing only a finite number of pairs $(x,y)$ such that $x$ divides $y$, then with no change about its density, we may assume that it contains no such pair. Using the above for $r=1$ and some classical evaluations of $\\pi (x)$, it leads to $\\lim \\frac {|S(x)|} x = 0$, so that $S$ cannot have a positive density.\r\n\r\nI hope it can be proved easier. Anyway, now we got the answer...\r\n\r\nPierre.", + "Solution_3": "[quote=\"pbornsztein\"]\n$ A(x) <\\pi (x) + c_2 \\frac {x^{\\frac 2 {r+1}}} {(\\ln x)^2} $\n\nwhere $A(x)$ is the number of terms not exceeding $x$.\n\n[/quote]\r\nThis is a bit strange because if this estimate were true for all $x$, we would have (for $r=1$) $A(x)\\leq C\\frac x{\\log x}$ while it is clear that after you construct several first terms in the sequence, say $a_1,\\dots,a_n$ (all greater than $1$), you can take a block from $N$ to $2N-1$ with very big $N$ and choose all numbers of the kind $k\\cdot a_1\\ldots a_n+1$ in that block. That will give you $A(2N-1)\\geq \\frac N{a_1\\dots a_n}$, which, for large $N$, will be much bigger than $N/\\log N$. And then you can do it again as many times as you wish... Pierre, do you mean that Erdos's upper estimate holds just for infinitely many $x$ rather than for all $x$? This is also enough to solve the problem. I'm also a bit perplexed with the lower estimate. What does it represent: existense of a set with that many elements for all $x$? :?", + "Solution_4": "I don't know....maybe is the evaluation only true for $r>1$?\r\nSomeone has to find the paper...\r\n\r\nPierre.", + "Solution_5": "What is absolutely remarkable is that Besikovitch constructed a set of positive upper density such that no two elements from it are divisible. On the other hand, it seems Erdos (of course) proved the assertion in the problem for the first time.", + "Solution_6": "[quote=\"harazi\"]Is it true that any set of positive density of natural numbers must contain infinitely many pairs $(x,y)$ of different elements such that x divides y? This started from a problem posted by Myth:if the sequence $(a_n)$ of natural numbers is increasing and $a_{n+1}-a_n$ is bounded, then this sequence has the described property. Does anyone know some results related to this problem?[/quote]\r\nActually, it was my question too when I was solving those problem.\r\nI don't understand what Harazi meant for \"upper positive density\". What is upper positive density? Is it different from positive density?\r\n\r\nSo do we know the answer for this problem?", + "Solution_7": "Probably it is a $\\limsup$ instead of a $\\lim$.\r\n\r\nPierre.", + "Solution_8": "Is density $\\lim$? Not $\\liminf$?", + "Solution_9": "Yes, Myth, we know the answer to these questions: for any set of positive density A there exists infinitely many couples a,b such that a divides b. THis result was proved by Erdos. On the other hand, Besicovitch constructed a set A such that the superiour limit of $ x_n/n$ is positive and there is no a,b in A such that a divides b. Here $x_n$ is the number of elements of A smaller than or equal to n. Now, it would be very nice to see both the proof of the initial problem and the construction of Besicovitch.", + "Solution_10": "Have you any references?\r\n\r\nPierre." +} +{ + "Tag": [ + "group theory", + "abstract algebra", + "algebra", + "function", + "domain", + "Ring Theory", + "superior algebra" + ], + "Problem": "How do I find the minimal subgroup of $ (\\mathbb Q, \\plus{})$ containing $ \\{\\frac{2}{3}, \\frac{3}{2}\\}$?", + "Solution_1": "It is $ \\{m\\cdot\\frac{2}{3}\\plus{}n\\cdot\\frac{3}{2}|m,n\\in\\mathbb{Z}\\}$.", + "Solution_2": "Switch to common denominator to see that the minimal subgroup $ U$ is given by:\r\n\r\n$ U\\equal{}\\frac{2}{3}\\cdot\\mathbb{Z}\\plus{}\\frac{3}{2}\\cdot\\mathbb{Z}\\equal{}\\frac{4\\cdot\\mathbb{Z}\\plus{}9\\cdot\\mathbb{Z}}{6}\\equal{}\\frac{1}{6}\\cdot\\mathbb{Z}$. \r\n\r\nAs for the last equality, note, that the inclusion \"$ \\subseteq$\" is obvious, and for the other inclusion use $ \\frac{n}{6}\\equal{}\\frac{4\\cdot (\\minus{}2n)\\plus{}9\\cdot n}{6}$ for abritrary $ n\\in\\mathbb{Z}$", + "Solution_3": "generalization: every finitely generated subgroup of $ \\mathbb{Q}$ (or the quotient field of any principal ideal domain) is cyclic, namely\r\n\r\n$ \\langle a_1 / b_1 , ... , a_n / b_n \\rangle \\equal{} \\langle gcd(a_1,...,a_n)/lcm(b_1,...,b_n) \\rangle$." +} +{ + "Tag": [ + "factorial", + "videos" + ], + "Problem": "Express $\\frac{10!-8!}{10!+8!}$ as a common fraction.", + "Solution_1": "This is a countdown...\r\n[hide]Factor out 8! to get $\\frac{90-1}{90+1}=\\frac{89}{91}$[/hide]", + "Solution_2": "I am assuming the use of no calculators.\r\n\r\n[hide]\nI factored out the $8!$ from the numerator and denominator. Then, it comes to $\\frac{8!(10\\cdot 9-1)}{8!(10\\cdot 9+1)}$. After that, the $8!$ cancel out and get $\\frac{90-1}{90+1}$. That is equal to $\\frac{89}{91}$.\n[/hide]", + "Solution_3": "You're both correct. :)\r\n\r\n@ bpms: It's a state sprint from 8 years ago.", + "Solution_4": "[quote=\"i_like_pie\"]You're both correct. :)\n\n@ bpms: It's a state sprint from 8 years ago.[/quote]\r\nAre you sure it isn't nationals? I could have sworn that I have seen this before on a nationals round, (perhaps 98...). What I meant is that this is a countdown level problem. I have seen variations of it myself at states when I was a 6th grader (2005) and I saw a variation on the video of the 2004 nationals video.", + "Solution_5": "[quote=\"bpms\"][quote=\"i_like_pie\"]You're both correct. :)\n\n@ bpms: It's a state sprint from 8 years ago.[/quote]\nAre you sure it isn't nationals? I could have sworn that I have seen this before on a nationals round, (perhaps 98...). What I meant is that this is a countdown level problem. I have seen variations of it myself at states when I was a 6th grader (2005) and I saw a variation on the video of the 2004 nationals video.[/quote]\r\nState Sprint 1998-99. I can PM you all the questions if you want...", + "Solution_6": "[quote=\"i_like_pie\"][quote=\"bpms\"][quote=\"i_like_pie\"]You're both correct. :)\n\n@ bpms: It's a state sprint from 8 years ago.[/quote]\nAre you sure it isn't nationals? I could have sworn that I have seen this before on a nationals round, (perhaps 98...). What I meant is that this is a countdown level problem. I have seen variations of it myself at states when I was a 6th grader (2005) and I saw a variation on the video of the 2004 nationals video.[/quote]\nState Sprint 1998-99. I can PM you all the questions if you want...[/quote]\r\nI have em somewhere...", + "Solution_7": "[quote=\"i_like_pie\"][quote=\"bpms\"][quote=\"i_like_pie\"]You're both correct. :)\n\n@ bpms: It's a state sprint from 8 years ago.[/quote]\nAre you sure it isn't nationals? I could have sworn that I have seen this before on a nationals round, (perhaps 98...). What I meant is that this is a countdown level problem. I have seen variations of it myself at states when I was a 6th grader (2005) and I saw a variation on the video of the 2004 nationals video.[/quote]\nState Sprint 1998-99. I can PM you all the questions if you want...[/quote]\r\n\r\nDo you have answers?", + "Solution_8": "[quote=\"anirudh\"][/quote][quote=\"i_like_pie\"][quote=\"bpms\"][quote=\"i_like_pie\"]You're both correct. :)\n\n@ bpms: It's a state sprint from 8 years ago.[/quote]\nAre you sure it isn't nationals? I could have sworn that I have seen this before on a nationals round, (perhaps 98...). What I meant is that this is a countdown level problem. I have seen variations of it myself at states when I was a 6th grader (2005) and I saw a variation on the video of the 2004 nationals video.[/quote]\nState Sprint 1998-99. I can PM you all the questions if you want...[/quote][quote=\"anirudh\"]\n\nDo you have answers?[/quote]\r\nFor the 15/30 that I did correctly. :blush:", + "Solution_9": "how do u remember all the questions and answers?\r\n-jorian", + "Solution_10": "[quote=\"jhredsox\"]how do u remember all the questions and answers?\n-jorian[/quote]\r\nHe probably has a copy of the test that his coach gave him, and he used in practice.", + "Solution_11": "[quote=\"jhredsox\"]how do u remember all the questions and answers?\n-jorian[/quote]\r\nI kept a copy. So does anybody want it? I can type it all up using LaTeX.", + "Solution_12": "a copy is on TZ's website :wink:", + "Solution_13": "[quote=\"i_like_pie\"]Express $\\frac{10!-8!}{10!+8!}$ as a common fraction.[/quote]\r\n[hide=\"solution\"]\n$\\frac{10!-8!}{10!+8!}$\n$\\frac{(90-1)8!}{(90+1)8!}$\n$\\frac{89}{91}$\n[/hide]", + "Solution_14": "[hide=\"solution\"]$\\frac{10!-8!}{10!+8!}= \\frac{8! \\times 89}{8! \\times 91}= \\frac{89}{91}$[/hide]", + "Solution_15": "[hide]\nIf you factor the $8!$ out, you get\n$8!\\frac{90-1}{90+1}=\\frac{89}{91}$[/hide]", + "Solution_16": "[hide] factor the 8! out... to make \n8!(90-1)/8!(90+1)...so that makes 89/91.\n[/hide]" +} +{ + "Tag": [], + "Problem": "For those of you who are interested in applying for Chennai Mathematical Institute:\r\nhttp://www.cmi.ac.in/admissions/\r\n\r\nGeneral Website:: http://cmi.ac.in", + "Solution_1": "thanks amigo :D \r\nFool chappli :mad:", + "Solution_2": "The exam is on May 29th. Who are all writing?", + "Solution_3": "when is the last date for submission of the form? I might chumma write as its after bits only :D", + "Solution_4": "i am i suppose", + "Solution_5": "[quote=\"pardesi\"]i am i suppose[/quote]\r\nbut dont you have direct admission being a KVPY scholar da??? :maybe:", + "Solution_6": "no that' only for IISER :)", + "Solution_7": "No. Only Students who have qualified to Maths Olympiad have direct admission. And Ironically Results of JEE are coming out on May 30th.", + "Solution_8": "again i ask, when's the last date for appcln of the form? :maybe:", + "Solution_9": "[quote=\"madness\"]again i ask, when's the last date for appcln of the form? :maybe:[/quote]\r\n\r\nAgain I tell you. They haven't really specified any such date. However atleast download it before end of march. :D", + "Solution_10": "they wouldnt wanna not provide applcn form to even the few that vouch to write the exam :P :rotfl:" +} +{ + "Tag": [ + "geometry proposed", + "geometry" + ], + "Problem": "I am very proud :) because I win the price for this problem. i want to post this but only now I could because now our official site put his solution so.\r\nLet a triangle $ABC$ with his angles $\\leq 90$. prove that there exist $(M_i,A_i,B_i,C_i)$ with $i=1,2$ such that:\r\ni)$M_i$ interior point \r\nii)$M_iA_i+M_iB_i+M_iC_i=2(M_iA+M_iB+M_iC)$\r\niii)The triangles $A_iBC,B_iCA,C_iAB$ are similar", + "Solution_1": "No solutions for long days. I remind to you that this problem exist", + "Solution_2": "After two months I will post my solution of this interesting problem which gave me the price, but because it is too long I will post in Greek, but it is easy to understand the solution.", + "Solution_3": "Silouan, \r\n\r\nplease see again the second condition in the problem. I think there is a typo there.\r\n\r\n[u]Babis[/u]", + "Solution_4": "I am so sorry. I edit the typo." +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "I'm interested in how could this problem be solved with lagrange multiplicators and similar methods:\r\n\r\n1.We have a,b,c,d,e real numbers such that a+b<=5 c+d+e<=5 0<=e<=d<=c<=b<=a Now find maximum value of P=a^2+b^2+c^2+d^2+e^2,and values of a,b,c,d,e \r\n when P is maximum.\r\n\r\nPlease if you know how to solve this problem.write it because elementary solution is quite stupid....", + "Solution_1": "You can solve these kind of problems using Convexity{Weighted Jensen's inequality}. \r\nI do not know Lagrange Multiplier's Method. So can't help you in that.", + "Solution_2": "[quote=\"Smilley\"]I'm interested in how could this problem be solved with lagrange multiplicators and similar methods:\n\n1.We have a,b,c,d,e real numbers such that a+b<=5 c+d+e<=5 0<=e<=d<=c<=b<=a Now find maximum value of P=a^2+b^2+c^2+d^2+e^2,and values of a,b,c,d,e \n when P is maximum.\n\nPlease if you know how to solve this problem.write it because elementary solution is quite stupid....[/quote]\r\nactually, i think we only use Lagrange Multipler when we have something like this:f(a,b,c,d,e)=k=const ,....\r\nso i think we can't use lagrange to solve this inequality :D", + "Solution_3": "[quote] You can solve these kind of problems using Convexity{Weighted Jensen's inequality}. [/quote]\r\nOk,could you write how , im very interested :?:", + "Solution_4": "See and here http://en.wikipedia.org/wiki/Lagrange_multipliers :wink:" +} +{ + "Tag": [ + "email" + ], + "Problem": "Do u have any siblings??", + "Solution_1": "Are there any siblings on the site? :) We know that there are a couple of parent-child couples, but any brother-sister, brother-brother, sister-sister? :)", + "Solution_2": "I've got two brothers and a baby sister. Sometimes I wish I wasn't the oldest.", + "Solution_3": "[quote=\"Valentin Vornicu\"]Are there any siblings on the site? :) We know that there are a couple of parent-child couples, but any brother-sister, brother-brother, sister-sister? :)[/quote]\r\n\r\nI tried to get my brother on this site, but he didn't want to. But he's only 8, so even a lot of the middle school problems would be hard for him.", + "Solution_4": "What were you doing trying to do that. Thats like child slavery! As much as I like to do math that would be unbarable.", + "Solution_5": "My parents used to not let me play outside if I didn't made some extra-math hours previously. At that point I hated that, but now I think I thank them :D", + "Solution_6": "no siblings :( wish i had some so my mom would have someone else to bother. does the number of siblings you have have anything to do with intelligence level?", + "Solution_7": "I have a one sister, she's older than me, but is still annoying. I'm probably annoying to her but anyways...\r\nshe's cool sometimes :D .", + "Solution_8": "i have an one and a half yr old sis. she can be annoying at times, but it's good to have someone to blame.... until my my parents announced \"from now on, you can blame her for everything, but YOU will have to clean the mess...\" \r\n!!!! arghhhhhh..... But she's still alsome to have as a sister.. right now she has noodles on her head!! :lol: hahaha...\r\nShe cracks us up so much, it's good to have her around.", + "Solution_9": "i have a younger brother who's 12 and a half. he's quite a character. he just went into the bathroom, sat down, and asked if one of us would get his book for him. :lol: current time is approaching 10 minutes. lol. he can be quite annoying tho...", + "Solution_10": "I have a brother thats a year older, and he is also on this site!", + "Solution_11": "[quote=\"peter\"]I have a brother thats a year older, and he is also on this site![/quote]Aha! First sibling pair on the site (first announced that is). :roll: :P", + "Solution_12": "[quote=\"Valentin Vornicu\"][quote=\"peter\"]I have a brother thats a year older, and he is also on this site![/quote]Aha! First sibling pair on the site (first announced that is). :roll: :P[/quote]\r\n\r\nMy sister has been on this site before.....hee hee....(she messed around with one of my posts once......even though she's only 1and a half....heehee.) :P", + "Solution_13": "[quote=\"KBabe\"]My sister has been on this site before.....hee hee....(she messed around with one of my posts once......even though she's only 1and a half....heehee.) :P[/quote]So in a decade or so, we should be expecting some more income from your house :D", + "Solution_14": "I think that JGeneson and AGeneson siblings too.", + "Solution_15": "after years of waiting, i finally got a sister... :lol:", + "Solution_16": "Alsome! Hope u have a great time having her around! (btw, how many yrs of waiting?...)", + "Solution_17": "My sister registered, then she looked a problem, died, and so she isn't here anymore.", + "Solution_18": "I have 15 siblings \r\n\r\n11 of them are boys\r\n3 of them are girls\r\n\r\nif you are wondering where the other one went, it is a she-man.\r\n\r\njust kidding.\r\n\r\nI have one sis", + "Solution_19": "2 siblings, one pretty good in maths but went into arts, the other an all rounder (=", + "Solution_20": "i have 0 sibblings :( . It's strange how if you don't have sibblings, you wish you had some and if you have sibblings, you wish that you were an only child.\r\n\r\nI have a friend who looks somewhat like me (almost like my twin) and he really likes math. Don't know if he's a member, though :?", + "Solution_21": "[quote=\"plokoon51\"]I have a friend who looks somewhat like me (almost like my twin) and he really likes math. Don't know if he's a member, though :?[/quote]You can email him to tell him about the site ;)", + "Solution_22": "[quote=\"furious\"]does the number of siblings you have have anything to do with intelligence level?[/quote]\r\n\r\nInterestingly, it does. The effect size is not large, but in carefully controlled studies there is a statistically significant influence of birth order and family size on IQ. The best birth order to be in is first-born, and the best family size is four children. The theory is that a first-born child starts out as an only child, getting lots of adult attention. Then when the later siblings come, the first-born child gets to practice explaining things to them. All of this is good for just a few points on a typical IQ test. But of course I am glad to know this, as the first-born child in a family of four children. :D \r\n\r\nA good reference book with a discussion of this issue, and much more, is \r\n\r\nN. J. Mackintosh (1998). IQ and Human Intelligence. Oxford: Oxford University Press. \r\n\r\nThat is a very good book, which I recommend to parents continually.", + "Solution_23": "yay!! first born, well... and last born.... yep, I'm an only child, and I'm enjoying it, even though it might be better if you can have a break from your parents sometimes.... I think the only bad thing about being second or third born is that you always have to live up to ur brother or sister if they are smart... but if you are the oldest but have a sibling that's one or two years younger, you feel sad or bad if they are smarter than you.", + "Solution_24": "[quote=\"tokenadult\"][quote=\"furious\"]does the number of siblings you have have anything to do with intelligence level?[/quote]\n\nInterestingly, it does. The effect size is not large, but in carefully controlled studies there is a statistically significant influence of birth order and family size on IQ. The best birth order to be in is first-born, and the best family size is four children.[/quote]\r\nWithout having any studies to back this up, my anecdotal impression is this.\r\nFirst-born children are more often in contact with adults and older children, and have to absorb advanced-for-their-age language by osmosis and guesswork. They are therefore likelier to have more complex vocabularies and syntax, which certainly promotes IQ score, but my impression is that they learn this somewhat more as abstraction or pattern-matching than a younger sibling would. On the other hand, their younger siblings, if close in age, will tend to learn to speak and read earlier, through absorption and imitation of the older siblings as they learn (e.g. close-in-age young siblings watching educational television together, or learning to read at roughly the same time). Because they have more time to learn language at a child's pace the younger siblings will tend to have vocabulary and language skills that are less adult or advanced, but with more of the emotional associations and underlying semantics.\r\n\r\nSo a first-born child would be more likely to be suited for, say, mathematics, and a younger (close in age) sibling might make a better writer or poet.\r\n\r\nJust some extrapolations from my and others' families. Usual caveats apply.", + "Solution_25": "that applies (wierdly) in my family too, although it is kind of surprising, because I like math but my sister does not at all. Seriously, she is only interested in Runescape and Manga and anything unreal. Kind of surprising given both my parents are mathematicians.", + "Solution_26": "wow, for the people who have large families 5+, what it is like having so many siblings?", + "Solution_27": "well... i have 2 sisters... one younger & one older.... so that's lyk a family of 5, rite?\r\n\r\ni know someone who has 5 kids in their family.... and the 2 child(a boy) LOVES math and science... he's like a GENIUS... and he, and his 2 younger sisters have all been to mathcounts.... the boy's a junior, younger sisters are a freshman and a 7th grader...", + "Solution_28": "I have a younger sister. She refuses to do math unless its homework. She's smart but not as interested in math as I am.", + "Solution_29": "[quote=\"JMI\"]I have a younger sister. She refuses to do math unless its homework. She's smart but not as interested in math as I am.[/quote]\r\nI am exactly the same :) :( :) :( :wacko: (hard to decide!)" +} +{ + "Tag": [], + "Problem": "Happy Thanks giving everyone. How was your thanks giving, and what are you thankful for?", + "Solution_1": "I'm thankful for thanksgiving. otherwise who knows what i would be doing in school right now", + "Solution_2": "im very thankful for WAN\r\n\r\nno i'm not kidding :roll: \r\n\r\n-jorian" +} +{ + "Tag": [], + "Problem": "The message is too small. Please make the message longer before submitting.", + "Solution_1": "Whats so special. You won abishop good job.", + "Solution_2": "Hmm... 1...fxg3 complicates things\r\nIf 2. hxg3 then 2...g4 and White is mated. If 2. fxg3 then Black takes on h2 threatening the bishop, and then ...g4, leading to the same thing as above.\r\nOnly way for White is to take the bishop, then give it back after Black takes on h2 w/ rh6 and White plays fxg\r\nOtherwise Black plays ...g4 and mates again.", + "Solution_3": "Exactly, so in all, black wins the bishop back, and has superior pawn structure. Not to mention he has his rooks in whites position. So black just turned the tables.", + "Solution_4": "Oh, and, by the way, get Blitzin. It's a lot easier to use than Dasher.\r\n\r\nWhite should have played g4. Black will have to lose material anyways." +} +{ + "Tag": [ + "function", + "calculus", + "integration", + "real analysis", + "real analysis theorems" + ], + "Problem": "I don't quite understand lower(upper) semicontinuity. Could someone explain this?", + "Solution_1": "Well a function $f: X\\to \\mathbb{N}$, where $X$ is some topological space, is called \"upper semicontinuous\" iff for every $n\\in \\mathbb{N}$ the set $f^{-1}(\\{m\\leq n\\})$ (the preimage of the set of all natural numbers which are at most $n$) is closed. This says, intutively, that the value of $f$ at the limit point of some sequence is at most as large as the values of $f$ at the points of the sequence.\r\n\r\nThere's no real need to take $\\mathbb{N}$, but I only saw the term in connection with $\\mathbb{N}$, so I took $\\mathbb{N}$ here.", + "Solution_2": "Since there's order implicit in the definition, an upper or lower semicontinous function must be a mapping into an ordered set. I'll assume that we're talking about real valued (or extended real-valued) functions.\r\n\r\nHere's one way to start. These definitions are from the book [i]Measure and Integral[/i] by Wheeden and Zygmund:\r\n\r\nLet $f$ be defined on $E$ and let $x_0$ be a limit point of $E$ which lies in $E$. Then we say that $f$ is upper semicontinuous at $x_0$ iff\r\n\\[ \\limsup_{x\\to x_0;x\\in E}f(x)\\le f(x_0) \\]\r\nSimilarly, $f$ is lower semicontinuus at $x_0$ iff\r\n\\[ \\liminf_{x\\to x_0;x\\in E}f(x)\\ge f(x_0) \\]\r\nFor finite values of $f(x_0),$ this could be made more specific as follows:\r\n\r\n$f\\text{ usc at }x_0\\iff\\,\\forall\\,\\epsilon>0,\\,\\exists\\,\\delta>0$ such that if $d(x,x_0)<\\delta$ then $f(x)0,\\,\\exists\\,\\delta>0$ such that if $d(x,x_0)<\\delta$ then $f(x)>f(x_0)-\\epsilon.$\r\n\r\nThe authors then define lsc or usc relative to $E$ as having this property for all limit points of $E$ in $E.$ The first theorem they prove is what Peter Scholze said:\r\n\r\n$f$ is usc on $E$ iff $\\forall\\,a,\\,\\{x\\in E: f(x)a\\}$ is open in $E.$\r\n\r\nWhere do such functions naturally arise? Consider a collection $f_{\\alpha}$ of continuous functions on $E.$ Let $f(x)=\\sup_{\\alpha}f_{\\alpha}(x).$ Then the set $\\{x: f(x)>a\\}=\\bigcup_{\\alpha}\\{x: f_{\\alpha}(x)>a\\}$ is a union of open sets, hence open. Thus the supremum of a set of continuous functions is lower semicontinuous. Similarly, the infinum of continuous functions is upper semicontinuous.\r\n\r\nConsider a function which takes on the value 1 at one point and is zero everywhere else. That function is upper semicontinuous but not lower semicontinuous. It can be realized as an infinum of a collection of continuous functions but not as a supremum of continuous functions.\r\n\r\nA function that is both upper and lower semicontinuous at the same point is continuous at that point.", + "Solution_3": "Thank you for the lengthy explanation.\r\n\r\n[quote=\"Kent Merryfield\"]Similarly, $f$ is lower semicontinuus at $x_0$ iff\n\\[ \\liminf_{x\\to x_0;x\\in E}f(x)\\ge f(x_0) \\]\n[/quote]\r\n\r\nHow do I visualize this expression? An example would be useful.\r\n\r\nCan I evalute to see if at $x_0$, where $f(x_0) = \\infty$, is lower/upper semicontinuous ?\r\n\r\nI was given a semicontinuous extended-real function defined as follows\r\n\\[ \\tilde{f}(x) = \\left \\{ \\begin{array}{ll} f(x) & \\quad ,\\mbox{if}\\ x \\in X \\\\ \\infty & \\quad ,\\mbox{otherwise} \\end{array} \\right., \\]\r\nwhere $X$ is nonempty closed convex set and $f$ is real-valued function. Is this $\\tilde{f}(x)$ lower semicontinuous function at all $x \\in X$? How about $x \\notin X$? Are they upper semicontinuous?\r\n\r\nIs it correct that if $X$ is nonempty convex set but not closed, then $\\tilde{f}(x)$ is not lower semicontinuous at all points residing on the boundary of $X$ (are they upper semicontinuous?) but lower semicontinuous everywhere in the interior of $X$?" +} +{ + "Tag": [ + "function", + "inequalities", + "real analysis", + "real analysis unsolved" + ], + "Problem": "if $f:I->(0,infinit)$ is a function such that $lnf$ is convex,then $f$ is convex. :)", + "Solution_1": "Is this problem trivial? :(", + "Solution_2": "Yes. It suffices to observe that from the AM-GM inequality we have $ f^{t}(x) f^{1-t}(y)\\leq tf(x)+(1-t)f(y)$." +} +{ + "Tag": [], + "Problem": "The trinomials $ a_{1}x^{2} \\plus{} 2b_{1}x \\plus{} c_{1}$ and $ a_{2}x^{2} \\plus{} 2b_{2}x \\plus{} c_{2}$ take positive value for each x in Real.\r\nProve that the trinomial $ a_{1}a_{2}x^{2} \\plus{} b_{1}b_{2}x \\plus{} c_{1}c_{2}$ takes also positive value for each x in Real.", + "Solution_1": "$ D_1\\equal{}4b^2_1\\minus{}4a_1c_1<0$\r\n\r\n$ a_1c_1>b_1^2$ (1)\r\n\r\n$ D_2\\equal{}4b_2^2\\minus{}4a_2c_2<0$\r\n\r\n$ a_2c_2>b_2^2$ (2)\r\n\r\n(1)*(2):\r\n$ a_1a_2c_1c_2>b_1^2b_2^2$\r\n\r\n\r\n$ D\\equal{}b_1^2b_2^2\\minus{}4a_1a_2c_1c_20$ and $ a_k$ satisfy the relation $ \\sum_{k\\equal{}0}^n a_k u_{n\\minus{}k}\\equal{}1$ for all sufficiently large $ n$. Then $ \\lim_{k\\to\\infty}a_k\\equal{}\\Bigl(\\sum_m u_m\\Bigr)^{\\minus{}1}$.", + "Solution_2": "Here is a direct solution:\r\n\r\nDividing the both sides of the given relation by $ n!$, we see that the Cauchy product of $ 1/n!$ and $ a_n / n!$ is always equal to 1.\r\n\r\nLet $ f(x) \\equal{} \\sum_{n\\equal{}0}^{\\infty} \\frac{a_n}{n!} x^n$ be the exponential generating function of $ a_n$, then we have $ f(x)e^x \\equal{} \\frac{1}{1\\minus{}x}$. So $ f(x) \\equal{} \\frac{e^{\\minus{}x}}{1\\minus{}x} \\equal{} \\sum_{n\\equal{}0}^{\\infty} \\left( \\sum_{k\\equal{}0}^{n} \\frac{(\\minus{}1)^{k}}{k!} \\right) x^n$.\r\n\r\nTherefore, $ \\lim_{n\\to\\infty} \\frac{a_n}{n!} \\equal{} \\sum_{k\\equal{}0}^{\\infty} \\frac{(\\minus{}1)^k}{k!} \\equal{} \\frac{1}{e}$." +} +{ + "Tag": [ + "ratio", + "geometry" + ], + "Problem": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20184 is the previous topic\r\n\r\nA 60 degree arc on circle O has the same length as a 45 degree arc of circle P. What is the ratio of the area of circle O to the area of circle P?\r\n\r\nBilly", + "Solution_1": "[quote=\"solafidefarms\"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20184 is the previous topic\n\nA 60 degree arc on circle O has the same length as a 45 degree arc of circle P. What is the ratio of the area of circle O to the area of circle P?\n\nBilly[/quote]\r\nEDIT:\r\n[hide]\n$\\frac{60}{360}C_1=\\frac{45}{360}C_2$\n\nCancel\n\nRatio Of Circumferences is $\\frac{3}{4}$\n\nRatio of areas is $\\frac{9}{16}$\n\n[/hide]", + "Solution_2": "[hide=\"my answer\"]9/16, as you can easily discover that the ratio of the circumferences is 3/4, and as area is in the second dimension, we just square the ratio of the circumferences.[/hide]" +} +{ + "Tag": [ + "geometry" + ], + "Problem": "How many square feet are in three square yards?", + "Solution_1": "We know that 3 feet equals a yard. So, a square yard has side lengths of 1 yard by 1 yard. This also means it is 3 feet 3 feet. To find the area we multiply length and width. Therefore a square yard has 9 square feet.", + "Solution_2": "Shentang, while you are correct that 1 sq.yd=9 sq.ft, the question asks how many sq.ft are in 3 sq.yds.\r\nThus, 9*3=[b] 27.[/b]" +} +{ + "Tag": [], + "Problem": "Solve the following equation:\r\n\\[ \\sqrt{ \\frac {x^2\\minus{}3x\\plus{}2} {x^2\\plus{}2x}} \\equal{} 1\\plus{}x \\]", + "Solution_1": "[quote=\"arqady\"]Solve the following equation:\n\\[ \\sqrt {\\frac {x^2 \\minus{} 3x \\plus{} 2} {x^2 \\plus{} 2x}} \\equal{} 1 \\plus{} x\n\\]\n[/quote]\r\n$ x \\neq 0;x \\neq \\minus{}2$\r\n<=>$ (\\sqrt {\\frac {x^2 \\minus{} 3x \\plus{} 2} {x^2 \\plus{} 2x}})^2 \\equal{} (1 \\plus{} x)^2$\r\n<=>$ x^2\\minus{}3x\\plus{}2\\equal{}(x^2\\plus{}2x\\plus{}1)(x^2\\plus{}2x)$\r\n<=>$ x^4\\plus{}4x^3\\plus{}4x^2\\plus{}5x\\minus{}2\\equal{}0$\r\n<=>$ (x^2\\plus{}3x\\minus{}1)(x^2\\plus{}x\\plus{}2)\\equal{}0$\r\n...........\r\nok! :P", + "Solution_2": "OK, but I think the following reasoning is better:\r\nLet $ x^2\\plus{}2x\\equal{}a$ and $ x\\minus{}2\\equal{}b.$ Then we obtain $ b^2\\plus{}b\\equal{}a^2\\plus{}a.$ :wink: :P" +} +{ + "Tag": [ + "real analysis", + "real analysis unsolved" + ], + "Problem": "Hi! I'm a new visitor to this forum. It's my first post...\r\nWho are the fonctions assure $f'(u(x))=f(x)$? for exemple, there is $x \\mapsto \\arctan{\\frac{x+a}{1-ax}}$ pour $x,a \\in \\R$ et $ax \\neq 1$.\r\nExcuse me to my english! it's very bad :blush: .", + "Solution_1": "What we have to find $f$ or $u$", + "Solution_2": "[quote=\"b\u00e9nabar\"]Hi! I'm a new visitor to this forum. It's my first post...\nWho are the fonctions assure $f'(u(x))=f(x)$? for exemple, there is $x \\mapsto \\arctan{\\frac{x+a}{1-ax}}$ pour $x,a \\in \\R$ et $ax \\neq 1$.\nExcuse me to my english! it's very bad :blush: .[/quote]\r\nYou can use French. Just make it clear." +} +{ + "Tag": [ + "geometry", + "analytic geometry", + "function", + "quadratics" + ], + "Problem": "A and B are the points of intersection of y=x^2-2x and y=2-3x, respectively.\r\n\r\nFind the coordinates of A and B, given that the x coordinate of B is greater than that of A. \r\n\r\nFor this do I solve the simultaneous equation? But after that where would the coordinates come from?", + "Solution_1": "set the two functions equal and you will have a quadratic in x. Solve for the two x's, then plug both back into either equation and you will have their respective y's. Each x and their respective y is a point of intersection.", + "Solution_2": "Do I solve it by substitution? How do I rearrange the equations?", + "Solution_3": "u set the two $y$ expressions equal; that is, $x^2-2x=2-3x$." +} +{ + "Tag": [ + "inequalities", + "function", + "quadratics" + ], + "Problem": "Let $a,b,c,$ and $k$ be nonnegative real numbers. Prove that\r\n\r\n\\[(a^2+k+1)(b^2+k+1)(c^2+k+1) \\geq (k+2)^2(ab+bc+ca+k-1)\\]\r\n\r\nSetting $k=1$ gives the APMO 2004 problem.", + "Solution_1": "[b]I dedicate the following proof to Vasc.[/b]\r\n\r\nFirst we prove the auxiliary inequality\r\n\r\n$1+2abc+a^{2}+b^{2}+c^{2}\\geq 2bc+2ca+2ab$.\r\n\r\nIn fact,\r\n\r\n$\\left( a+b+c\\right) \\left( \\left( 1+2abc+a^{2}+b^{2}+c^{2}\\right)-\\left(2bc+2ca+2ab\\right) \\right)$\r\n$=\\left( \\left( a^{3}+b^{3}+c^{3}+3abc\\right)-\\left(bc^{2}+cb^{2}+ca^{2}+ac^{2}+ab^{2}+ba^{2}\\right) \\right)$\r\n$+a\\left( bc^{2}+cb^{2}+1-3bc\\right)+b\\left( ca^{2}+ac^{2}+1-3ca\\right)+c\\left( ab^{2}+ba^{2}+1-3ab\\right)$.\r\n\r\nNow, since Schur yields $a^{3}+b^{3}+c^{3}+3abc \\geq bc^{2}+cb^{2}+ca^{2}+ac^{2}+ab^{2}+ba^{2}$, we have $\\left( a^{3}+b^{3}+c^{3}+3abc\\right)-\\left(bc^{2}+cb^{2}+ca^{2}+ac^{2}+ab^{2}+ba^{2}\\right) \\geq 0$, and furthermore, AM-GM for the three numbers $bc^{2}$, $cb^{2}$ and 1 gives $bc^{2}+cb^{2}+1 \\geq 3\\sqrt[3]{bc^{2}\\cdot cb^{2}\\cdot 1}= 3bc$ and thus $bc^{2}+cb^{2}+1-3bc\\geq 0$ and similarly $ca^{2}+ac^{2}+1-3ca\\geq 0$ and $ab^{2}+ba^{2}+1-3ab\\geq 0$. Hence, we have\r\n\r\n$\\left( a+b+c\\right) \\left( \\left( 1+2abc+a^{2}+b^{2}+c^{2}\\right)-\\left(2bc+2ca+2ab\\right) \\right) \\geq 0$\r\n\r\nand thus $1+2abc+a^{2}+b^{2}+c^{2}\\geq 2bc+2ca+2ab$. This proves our auxiliary inequality. By the way, is there any nice proof?\r\n\r\n[b]EDIT:[/b] Yes, of course! \r\n\r\nAccording to the pigeonhole principle, among the three numbers a - 1, b - 1, c - 1, at least two have the same sign; WLOG, say that the numbers b - 1 and c - 1 have the same sign, so that $\\left(b-1\\right)\\left(c-1\\right)\\geq 0$. Then, according to the inequality $x^{2}+y^{2}\\geq 2xy$ for any two reals x and y, we have\r\n\r\n$\\left(b-1\\right)^{2}+\\left(c-1\\right)^{2}\\geq 2\\left(b-1\\right)\\left(c-1\\right)\\geq-2\\left(a-1\\right)\\left(b-1\\right)\\left(c-1\\right)$,\r\n\r\nwhere the last $\\geq$ sign is because $\\left(b-1\\right)\\left(c-1\\right)\\geq 0$ and $1\\geq-\\left(a-1\\right)$ (the latter is clear since $1\\geq 1-a$, since $a\\geq 0$). Thus,\r\n\r\n$\\left(1+2abc+a^{2}+b^{2}+c^{2}\\right)-\\left(2bc+2ca+2ab\\right)$\r\n$=\\left(a-1\\right)^{2}+\\left(b-1\\right)^{2}+\\left(c-1\\right)^{2}+2\\left(a-1\\right)\\left(b-1\\right)\\left(c-1\\right)$\r\n$\\geq\\left(a-1\\right)^{2}+\\left(-2\\left(a-1\\right)\\left(b-1\\right)\\left(c-1\\right)\\right)+2\\left(a-1\\right)\\left(b-1\\right)\\left(c-1\\right)$\r\n$=\\left(a-1\\right)^{2}\\geq 0$,\r\n\r\nand hence $1+2abc+a^{2}+b^{2}+c^{2}\\geq 2bc+2ca+2ab$.\r\n\r\n[b]EDIT EDIT:[/b] I see this is equivalent to Zhaobin's proof below. Anyway, I wanted to note that there is a weaker inequality by Harazi, Titu Andreescu and Zuming Feng: $a^{2}+b^{2}+c^{2}+2abc+3\\geq\\left(1+a\\right)\\left(1+b\\right)\\left(1+c\\right)$.\r\n\r\nNow, to the proof of our inequality:\r\n\r\n$\\left( a^{2}+k+1\\right) \\left( b^{2}+k+1\\right) \\left( c^{2}+k+1\\right)-\\left( k+2\\right)^{2}\\left( ab+bc+ca+k-1\\right)$\r\n$=\\frac12\\left( \\left( b-c\\right)^{2}+\\left( c-a\\right)^{2}+\\left( a-b\\right)^{2}\\right) k^{2}$\r\n$+\\left( \\left( b-c\\right)^{2}+\\left( c-a\\right)^{2}+\\left( a-b\\right)^{2}+\\left( bc-1\\right)^{2}+\\left( ca-1\\right)^{2}+\\left( ab-1\\right)^{2}\\right) k$\r\n$+\\left( bc-1\\right)^{2}+\\left( ca-1\\right)^{2}+\\left( ab-1\\right)^{2}+\\left( abc-1\\right)^{2}$\r\n$+\\left( \\left( 1+2abc+a^{2}+b^{2}+c^{2}\\right)-\\left( 2bc+2ca+2ab\\right)\\right)$.\r\n\r\nHereby, the last bracket is $\\geq 0$ after our auxiliary inequality, and all the other brackets are trivially $\\geq 0$. Hence,\r\n\r\n$\\left( a^{2}+k+1\\right) \\left( b^{2}+k+1\\right) \\left( c^{2}+k+1\\right)-\\left( k+2\\right)^{2}\\left( ab+bc+ca+k-1\\right) \\geq 0$,\r\n\r\nand\r\n\r\n$\\left( a^{2}+k+1\\right) \\left( b^{2}+k+1\\right) \\left( c^{2}+k+1\\right) \\geq \\left( k+2\\right)^{2}\\left( ab+bc+ca+k-1\\right)$.\r\n\r\nQED.\r\n\r\n[b]EDIT:[/b] See also http://www.mathlinks.ro/viewtopic.php?t=24093 for proofs of the auxiliary inequality.\r\n\r\n Darij", + "Solution_2": "Darij, this is really a very nice solution.\r\n\r\nCongratulations!!!\r\n\r\nHow did you think the trick of multiplication by a+b+c to prove the lemma?\r\n\r\nIt is very ingenious ;)", + "Solution_3": "Thanks, Manlio. Just to give an idiot answer to your question:\r\n\r\n[quote=\"manlio\"]How did you think the trick of multiplication by a+b+c to prove the lemma?[/quote]\r\n\r\nI tried everything possible with my computer algebra system and this was one of the variants I had tested. Although, I think, there should be a much simpler proof of $1+2abc+a^2+b^2+c^2 \\geq 2bc+2ca+2ab$.\r\n\r\n Darij", + "Solution_4": "Darij, did you complete the squares by hand? Or with the computer? Either way, how did you manage to do it?", + "Solution_5": "[quote=\"darij grinberg\"]Although, I think, there should be a much simpler proof of $1+2abc+a^2+b^2+c^2 \\geq 2bc+2ca+2ab$.[/quote]\r\n\r\nWell, there is a very straightforward (and more practical, contest-wise) way to prove this:\r\n\r\nconsider the function $1+2abc+a^2+b^2+c^2 -( 2bc+2ca+2ab)$ as a quadratic in $a$, then it being non-negative is equivalent to its discriminant being negative. \r\n\r\nThe discriminant is an function in $b, c$. The power of $b$ is at most 2, so we check again the discriminant in $b$, then we are left with a single variable function to analyse, which is easy in this case.", + "Solution_6": "[quote=\"darij grinberg\"]Although, I think, there should be a much simpler proof of $1+2abc+a^2+b^2+c^2 \\geq 2bc+2ca+2ab$.[/quote]\r\n\r\n\r\nWell in fact, there is one:\r\n\r\nFor convenience, set $x^3=a, y^3=b, z^3=c,$ so we have\r\n\r\n$x^6+y^6+z^6+x^3y^3z^3+x^3y^3z^3+1 \\geq x^6+y^6+z^6+3x^2y^2z^2$\r\n\r\n$\\geq \\sum_{\\textrm{cyclic}} x^4y^2+x^2y^4 \\geq \\sum_{\\textrm{cyclic}} 2x^3y^3$\r\n\r\n$=2(x^3y^3+y^3z^3+z^3x^3)$\r\n\r\nwhich follows from AM-GM + Schur + AM-GM.\r\n\r\n[edit] Well, I just realized that this isn't that much simpler... oh well.", + "Solution_7": "Dear Billzhao,\r\n\r\nI like your idea\r\n\r\n[quote=\"billzhao\"][quote=\"darij grinberg\"]Although, I think, there should be a much simpler proof of $1+2abc+a^2+b^2+c^2 \\geq 2bc+2ca+2ab$.[/quote]\n\nWell, there is a very straightforward (and more practical, contest-wise) way to prove this:\n\nconsider the function $1+2abc+a^2+b^2+c^2 -( 2bc+2ca+2ab)$ as a quadratic in $a$, then it being non-negative is equivalent to its discriminant being negative. \n\nThe discriminant is an function in $b, c$. The power of $b$ is at most 2, so we check again the discriminant in $b$, then we are left with a single variable function to analyse, which is easy in this case.[/quote]\r\n\r\nbut I cannot use it.\r\n\r\nIn fact, I have considered $f(a)= a^2+b^2+c^2 +2abc +1 -(2ab+2bc+2ca)= a^2+2a(bc-b-c) +b^2+c^2+1-2bc$\r\n\r\nWe have to cases to consider:\r\n\r\n1) if $bc \\geq b+c $ and it is obvious\r\n\r\n2) if $bc =GM\r\n\r\nsorry for my very very poor English", + "Solution_14": "another nice proof to a 2 +b 2 +c 2 +2abc+1 >= 2(ab+bc+ca)\r\n\r\nfor a 2 +b 2 +c 2 +2abc+1-2(ab+bc+ca)=(a 2 +1)+(b 2 +c 2)+2abc-2(ab+bc+ca) >= 2a+2bc+2abc-2ab-2bc-2ca=2a+2abc-2ab-2ca =2a(1-b)(1-c)\r\nfor there must two numbers of (1-a),(1-b),(1-c) have the same sign.we can assume (1-b)(1-c) >= 0 over :)", + "Solution_15": "[quote=\"ThAzN1\"]Let $a,b,c,$ and $k$ be nonnegative real numbers. Prove that\n\n\\[ (a^2+k+1)(b^2+k+1)(c^2+k+1) \\geq (k+2)^2(ab+bc+ca+k-1) \\]\n\nSetting $k=1$ gives the APMO 2004 problem.[/quote]\r\n [b]Thank you for the following problem.Yes, you are right.I omitted that<1>, but in essence that's the problem.Anywhay, I've known it.It was very good.I love inequalities.I've composed some problems whith inequalities.If you want, I can send you some of them(on your site). :) [/b]", + "Solution_16": "[quote=\"darij grinberg\"]\nFirst we prove the auxiliary inequality\n\n$1+2abc+a^{2}+b^{2}+c^{2}\\geq 2bc+2ca+2ab$.\n\n[/quote]\r\n\r\nI proved it by another ways.\r\n\r\n[b]Solution 1[/b]\r\nThe inequality can be obtained by the sum of the followings: \\[a^{2}+b^{2}+c^{2}-2bc-2ca-2ab+3(abc)^{\\frac{2}{3}}\\geq 0\\] by Schur inequality and AM-GM, and \\[2abc-3(abc)^{\\frac{2}{3}}+1 \\geq 0\\] which is equivalent to $2t^{3}-3t^{2}+1 \\geq 0 \\iff (t-1)^{2}(2t+1)\\geq 0$ where $t^{3}=abc$.\r\n\r\n[b]Solution 2[/b]\r\nThis is little bit messy, but I think this method is meaningful. This solution used homogenization in somewhat other sense, and then get symmetric homogenous inequality on $a,b,c$ through derivative.\r\n\r\nFirst let \\[f(x)=x^{3}+(\\sum a^{2}-2 \\sum ab )x+2abc.\\] I will show that for any positive real number $x$, $f(x) \\geq 0$. Then \\[a^{2}+b^{2}+c^{2}+2abc+1-2ab-2bc-2ca=f(1)\\geq0,\\] and the proof is completed. \r\n\r\nNote that $f'(x)=3x^{2}+\\sum a^{2}-2 \\sum ab$. \r\n\r\nCase 1) If $\\sum a^{2}-2 \\sum ab >0$, then $f'(x)>0$. So $f(x)>f(0)=2abc>0$.\r\nCase 2) Otherwise, $\\sum a^{2}-2 \\sum ab \\leq 0$. Then the minimum of $f(x)$ in $(0,\\inf)$ takes place at $x \\rightarrow 0$ or $x=\\sqrt{-\\frac{A}{3}}$ where $A=\\sum a^{2}-2 \\sum ab$. As we know that $f(0)=2abc>0$, we gotta check if $f(x_{0}) \\geq 0$, where $x_{0}=\\sqrt{-\\frac{A}{3}}$. \\[f(x_{0})=x_{0}^{3}+AX_{0}+2abc=\\frac{2}{3}Ax_{0}+2abc \\geq 0\\] \\[\\iff 3abc \\geq-\\frac{A}{3}\\cdot \\sqrt{-\\frac{A}{3}}\\] \\[\\iff 27a^{2}b^{2}c^{2}\\geq-A^{3}.\\] Hereby note that $A\\leq 0$ means that $\\sqrt{a}, \\sqrt{b}, \\sqrt{c}$ are the sides of a triangle. So there is $p,q,r$ such that $\\sqrt{a}=q+r, \\sqrt{b}=r+p, \\sqrt{c}=p+q$. \\[\\iff 27(p+q)^{4}(q+r)^{4}(r+p)^{4}\\geq 2^{12}(p+q+r)^{3}p^{3}q^{3}r^{3}\\] \\[\\iff (9uv^{2}-w^{3})^{4}\\geq 2^{12}u^{3}w^{9}\\] where $u= \\frac{a+b+c}{3}, v=\\sqrt{\\frac{ab+bc+ca}{3}}, w=\\sqrt[3]{abc}$.\r\n\r\nNote that $u \\geq v \\geq w$. The last inequality is true as $(9uv^{2}-w^{3})^{4}\\geq (8uv^{2})^{4}= 2^{12}u^{4}v^{8}= 2^{12}u^{3}(uv^{8}) \\geq 2^{12}u^{3}w^{9}$. This means that $f(x_{0}) \\geq 0$, which completes the proof.", + "Solution_17": "This inequality was given at a Romanian contest.I solved it like Darij's solution 2.", + "Solution_18": "[quote=\"Sung-yoon Kim\"][quote=\"darij grinberg\"]\nFirst we prove the auxiliary inequality\n\n$1+2abc+a^{2}+b^{2}+c^{2}\\geq 2bc+2ca+2ab$.\n\n[/quote]\n\n[b]Solution 2[/b] \\[27(p+q)^{4}(q+r)^{4}(r+p)^{4}\\geq 2^{12}(p+q+r)^{3}p^{3}q^{3}r^{3}\\] [/quote]\r\n\r\nMoreover, these two inequalities are EQUIVALENT.", + "Solution_19": "Coming back to the generalization, I have another solution. Actually I applied $(*)$ for the original APMO 2004 inequality, and I noticed it may be applied as well to this generalization. I will prove the following:\r\n\\[(a^{2}+k)(b^{2}+k)(c^{2}+k)\\geq(k+1)^{2}\\left(\\frac{(a+b+c)^{2}}{3}+k-2\\right). \\]\r\nfor $k\\geq1$ and $a,b,c\\geq0$.\r\n\r\n$(*)$ The main idea is to get rid of the $a^{2}b^{2}c^{2}$ which appears when oppening the brakets. This can be done in the following way: $a^{2}b^{2}c^{2}+a^{2}\\geq 2(ab)(bc)$. \r\nThe rest of the terms of the RHS can be easily minimized to a function of $t=ab+bc+ca$ using $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\\geq\\frac{t^{2}}{3}$ and $s=a^{2}+b^{2}+c^{2}\\geq t$.\r\n\r\nSo, $RHS=a^{2}b^{2}c^{2}+k(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+k^{2}(a^{2}+b^{2}+c^{2})+k^{3}$ $=\\left[a^{2}b^{2}c^{2}+\\frac{a^{2}+b^{2}+c^{2}}{3}\\right]+$ $k(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+\\left(k^{2}-\\frac{1}{3}\\right)(a^{2}+b^{2}+c^{2})+k^{3}$ which is $\\geq$ than\r\n$\\frac{1}{3}\\left(2a^{2}bc+2b^{2}ac+2c^{2}ab+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\\right)+\\left(k-\\frac{1}{3}\\right)(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$ $+\\left(k^{2}-\\frac{1}{3}\\right)(a^{2}+b^{2}+c^{2})+k^{3}.$\r\n\r\nSo, \r\n$RHS\\geq\\frac{1}{3}t^{2}+\\left(k-\\frac{1}{3}\\right)(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$ $+\\left(k^{2}-\\frac{1}{3}\\right)s+k^{3}\\geq$\r\n$\\frac{t^{2}}{3}+\\left(k-\\frac{1}{3}\\right)\\frac{t^{2}}{3}+\\left(k^{2}-\\frac{1}{3}\\right)s+k^{3}=$\r\n$E=\\left(\\frac{3k+2}{3}\\right)\\frac{t^{2}}{3}+\\left(k^{2}-\\frac{1}{3}\\right)s+k^{3}$.\r\n\r\nNow, $LHS=\\frac{(k+1)^{2}}{3}s+\\frac{2(k+1)^{2}}{3}t+k^{3}-(3k+2)$.\r\n\r\nNow, $E-LHS=\\left(\\frac{3k+2}{3}\\right)\\frac{t^{2}}{3}$ $+\\left(k^{2}-\\frac{1}{3}-\\frac{(k+1)^{2}}{3}\\right)s$ $-\\frac{2(k+1)^{2}}{3}t+(3k+2)$.\r\nBecause $s\\geq t$ and $k^{2}-\\frac{1}{3}-\\frac{(k+1)^{2}}{3}\\geq0$, as $k\\geq1$, we get \r\n$E-LHS\\geq (3k+2)\\left(\\frac{t^{2}}{9}-\\frac{2}{3}t+1\\right)=(3k+2)\\left(\\frac{t}{3}-1\\right)^{2}\\geq0$.\r\n\r\nMaybe the above seems long but it's not complicated. After you use $(*)$ all calculations are obvious, or at least straighforward.", + "Solution_20": "[quote=\"darij grinberg\"]Although, I think, there should be a much simpler proof of $ 1+2abc+a^{2}+b^{2}+c^{2}\\geq 2bc+2ca+2ab$.[/quote]\r\n\r\nsee\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=155443\r\n\r\ncarij\r\n\r\n[Darij edit: :rotfl: ]", + "Solution_21": "Thank you very much.", + "Solution_22": "[quote=\"darij grinberg\"]\n\n$ 1 \\plus{} 2abc \\plus{} a^{2} \\plus{} b^{2} \\plus{} c^{2}\\geq 2bc \\plus{} 2ca \\plus{} 2ab$.\n\n[/quote]\r\n\r\nYou can also prove it this way:\r\nLet $ k \\equal{} \\sqrt[3]{abc}$\r\nThen by Schur and AM-GM we have:\r\n$ a^2\\plus{}b^2\\plus{}c^2\\plus{}3k^2 \\geq 2ab\\plus{}2bc\\plus{}2ca$\r\nAlso by AM-GM:\r\n$ abc \\plus{} abc \\plus{} 1 \\geq 3k^2$\r\nAdding those two inequalities yields Darij's auxiliary inequality", + "Solution_23": "[quote=\"darij grinberg\"][b]I dedicate the following proof to Vasc.[/b]\n\nFirst we prove the auxiliary inequality\n\n$ 1 \\plus{} 2abc \\plus{} a^{2} \\plus{} b^{2} \\plus{} c^{2}\\geq 2bc \\plus{} 2ca \\plus{} 2ab$.[/quote]\r\n\r\nWe have also the following generalization for $ 0\\le k \\le 1$:\r\n\r\n$ a^2\\plus{}b^2\\plus{}c^2\\plus{}2kabc\\plus{}k \\ge (k\\plus{}1)(ab\\plus{}bc\\plus{}ca)$", + "Solution_24": "Very nice one, Vasc :wink: Here is my solution for this.\r\nWe have:\r\n\\[ f(a,b,c,k)\\equal{}a^2\\plus{}b^2\\plus{}c^2\\plus{}2kabc\\plus{}k\\minus{}(k\\plus{}1)(ab\\plus{}bc\\plus{}ca)\\]\r\nWe can assume that $ a\\le 1$\r\nIt is easy to see that:\r\n\\[ f(a,b,c,k)\\minus{}f\\left(a,\\frac{b\\plus{}c}{2},\\frac{b\\plus{}c}{2},k\\right)\\equal{}(b\\minus{}c)^2(3\\plus{}k\\minus{}2ak)\\ge 0\\]\r\n\r\nHence we have:\r\n\\[ f\\left(a,\\frac{b\\plus{}c}{2},\\frac{b\\plus{}c}{2},k\\right)\\equal{}f(a,t,t,k)\\]\r\n\\[ \\equal{}(2ak\\minus{}k\\plus{}1)t^2\\minus{}2(ka\\plus{}2a)t\\plus{}a^2\\plus{}k\\ge 0,\\forall k\\in [0,1]\\]\r\nIn fact:\r\n\\[ \\Delta' \\equal{}(ka\\plus{}2a)^2\\minus{}(2ak\\minus{}k\\plus{}1)(a^2\\plus{}k)\\equal{}\\minus{}k(a\\minus{}1)^2(1\\minus{}k\\plus{}2a)< 0\\]\r\nWe are done.", + "Solution_25": "[b]Vasc wrote:[/b]\r\n[quote]We have also the following generalization for $ 0\\leq k\\leq 1$ : \n$ a^2\\plus{}b^2\\plus{}c^2\\plus{}2kabc\\plus{}k\\geq (k\\plus{}1)(ab\\plus{}bc\\plus{}ca)$.[/quote]\r\nWhat is your proof for this one, Vasc?", + "Solution_26": "[quote=\"zaizai-hoang\"]Very nice one, Vasc :wink: Here is my solution for this.\nWe have:\n\\[ f(a,b,c,k) \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 2kabc \\plus{} k \\minus{} (k \\plus{} 1)(ab \\plus{} bc \\plus{} ca)\n\\]\nWe can assume that $ a\\le 1$\nIt is easy to see that:\n\\[ f(a,b,c,k) \\minus{} f\\left(a,\\frac {b \\plus{} c}{2},\\frac {b \\plus{} c}{2},k\\right) \\equal{} (b \\minus{} c)^2(3 \\plus{} k \\minus{} 2ak)\\ge 0\n\\]\nHence we have:\n\\[ f\\left(a,\\frac {b \\plus{} c}{2},\\frac {b \\plus{} c}{2},k\\right) \\equal{} f(a,t,t,k)\n\\]\n\n\\[ \\equal{} (2ak \\minus{} k \\plus{} 1)t^2 \\minus{} 2(ka \\plus{} 2a)t \\plus{} a^2 \\plus{} k\\ge 0,\\forall k\\in [0,1]\n\\]\nIn fact:\n\\[ \\Delta' \\equal{} (ka \\plus{} 2a)^2 \\minus{} (2ak \\minus{} k \\plus{} 1)(a^2 \\plus{} k) \\equal{} \\minus{} k(a \\minus{} 1)^2(1 \\minus{} k \\plus{} 2a) < 0\n\\]\nWe are done.[/quote]\r\n\r\nCan you please explain to me why we can assume that $ a\\le 1$ ?\r\n\r\nThank you very much. :)", + "Solution_27": "[quote=\"Inequalities Master\"][b]Vasc wrote:[/b]\n[quote]We have also the following generalization for $ 0\\leq k\\leq 1$ : \n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 2kabc \\plus{} k\\geq (k \\plus{} 1)(ab \\plus{} bc \\plus{} ca)$.[/quote]\nWhat is your proof for this one, Vasc?[/quote]\r\nHint: It is a monotonic function in terms of $ k$, hence it suffices to consider the case $ k\\equal{}0$ and $ k\\equal{}1$ (but in these cases, it is very easy to prove) :)", + "Solution_28": "[quote=\"darij grinberg\"]\nI tried everything possible with my computer algebra system and this was one of the variants I had tested. Although, I think, there should be a much simpler proof of $ 1 \\plus{} 2abc \\plus{} a^2 \\plus{} b^2 \\plus{} c^2 \\geq 2bc \\plus{} 2ca \\plus{} 2ab$.\n Darij[/quote]\nI think, the simplest proof is :\nAssume that $ (a\\minus{}1)(b\\minus{}1) \\ge 0$, then\n\\[ 1 \\plus{} 2abc \\plus{} a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} 2(bc\\plus{}ca\\plus{}ab)\\equal{}(a\\minus{}b)^2\\plus{}(c\\minus{}1)^2\\plus{}2c(a\\minus{}1)(b\\minus{}1) \\ge 0\\]\n\n[quote=\"Vasc\"] \nWe have also the following generalization for $ 0\\le k \\le 1$:\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 2kabc \\plus{} k \\ge (k \\plus{} 1)(ab \\plus{} bc \\plus{} ca)$[/quote]\r\nThis is true because \r\n$ LHS\\minus{}RHS\\equal{}(1\\minus{}k)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab\\minus{}bc\\minus{}ca)\\plus{}k(a\\minus{}b)^2\\plus{}k(c\\minus{}1)^2\\plus{}2kc(a\\minus{}1)(b\\minus{}1) \\ge 0$ \r\nwhere we assume $ (a\\minus{}1)(b\\minus{}1) \\ge 0$.\r\n\r\nBy the same way, we can prove the following inequality for all $ a,b,c >0$: \r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} pabc \\plus{} \\frac{4(q\\minus{}1)^2}{p^2} \\ge q(ab \\plus{} bc \\plus{} ca)$ \r\nwhere $ p>0$ and $ 1\\le q \\le 2$.\r\nMoreover, if we denote $ P(a,b,c)\\equal{}a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} pabc \\minus{}q(ab \\plus{} bc \\plus{} ca)$ for all $ a,b,c>0$, then:\r\n+ if $ p<0$ or $ p\\equal{}0,q>1$ or $ p>0,q>2$ then $ \\inf P\\equal{}\\minus{}\\infty$\r\n+ if $ p\\ge 0,q\\le 1$ then $ \\min P\\equal{}0$.", + "Solution_29": "[quote=\"ehku\"] \nBy the same way, we can prove the following inequality for all $ a,b,c > 0$: \n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} pabc \\plus{} \\frac {4(q \\minus{} 1)^2}{p^2} \\ge q(ab \\plus{} bc \\plus{} ca)$ \nwhere $ p > 0$ and $ 1\\le q \\le 2$.\n[/quote]\r\nI have found that this inequality is equivalent to the above Vasc's one.", + "Solution_30": "[quote=\"Vasc\"]Find all reals $ k_1$ and $ k_2$ and $ k_3$ such that\n\n $ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq k_1(bc + ca + ab)^2$ (1)\n\nand\n\n $ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq k_2(a + b + c)^2$ (2)\n\nand\n\n $ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq k_3(a + b + c)$ (3)\n\nhold for all $ a$, $ b$ and $ c$ real numbers.[/quote]$ \\prod{(a^2 + 1)} - \\frac {3}{4}(\\sum{bc})^2$\n\n$ = \\frac {1}{6}\\sum{(b + c - abc)^2} + \\frac {1}{6}\\sum{a^2(bc - 2)^2} + \\frac {1}{48}\\sum{(ca + ab - 4)^2}$\n\n$ + \\frac {5}{48}\\sum{a^2(b - c)^2}\\geq0\\Longrightarrow$\n\n$ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq\\frac {3}{4}(bc + ca + ab)^2,$ \n\nwith equality if and only if $ a = b = c = \\pm\\sqrt {2}.$\n\n\n$ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq\\frac {3}{4}(a + b + c)^2,$ \n\nwith equality if and only if $ a = b = c = \\pm\\frac {1}{\\sqrt {2}}.$\n\nSee here : http://www.mathlinks.ro/viewtopic.php?t=274139\n\n\n$ \\prod{(a^2 + 1)} - \\frac {72}{25\\sqrt {5}}\\sum{a}$\n\n$ = \\frac {12}{25}(\\frac {3}{\\sqrt {5}} - \\sum{a})^2 + \\frac {17}{45}(\\frac {3}{5} - \\sum{bc})^2 + \\frac {1}{3}\\sum{a^2(bc - \\frac {1}{5})^2}$\n\n$ + \\sum{(b - c)^2(\\frac {14}{45}a^2 + \\frac {19}{75})}\\geq0\\Longrightarrow$\n\n$ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq\\frac {72}{25\\sqrt {5}}(a + b + c),$ \n\nwith equality if and only if $ a = b = c = \\frac {1}{\\sqrt {5}}.$\n[quote=\"ThAzN1\"]$ (a^2 + k + 1)(b^2 + k + 1)(c^2 + k + 1) \\geq (k + 2)^2( bc + ca + ab + k - 1).$[/quote]This inequality holds for all real numbers $ a,b,c$ if and only if $ k\\geq - \\frac {2}{3}.$\r\n\r\nA proof see : http://www.mathlinks.ro/viewtopic.php?t=257228", + "Solution_31": "Let $a,b,c> 0. $ Prove that\n$$ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq\\frac {72}{25\\sqrt 5}(a + b + c)$$ \n[quote=Ji Chen]\n$$ (a^2 + 1)(b^2 + 1)(c^2 + 1) \\geq\\frac {3}{4}(a + b + c)^2$$ \n[/quote]\n[url]https://artofproblemsolving.com/community/c6h1263156p13551684[/url]\n Let $a,b,c >0.$ [url=http://www.artofproblemsolving.com/community/c6h1133617p5280576]Prove that[/url] $$ \\left (a ^{2}+1\\right )\\left ( b^{2}+1 \\right )\\left ( c^{2}+1 \\right )\\geq \\frac{5}{16}\\left ( a+b+c+1 \\right )^{2} $$", + "Solution_32": "[quote=sqing] Let a,b,c >0. Prove $ \\left (a ^{2}+1\\right )\\left ( b^{2}+1 \\right )\\left ( c^{2}+1 \\right )\\geq \\frac{5}{16}\\left ( a+b+c+1 \\right )^{2} $\n[/quote]\nThe following general one is also true:\nLet $a,b,c\\geq 0$ and $k\\geq 0$, then\n\\[(a^2+1)(b^2+1)(c^2+1)\\geq \\alpha_k(a+b+c+k)^2\\]\nwhere $\\alpha_k=\\frac{(k^2+12-k\\sqrt{k^2+8})^3}{32(3\\sqrt{k^2+8}+k)^2}$. This can be proved by CS with some simple calculation.", + "Solution_33": "[quote=ThAzN1]Let $a,b,c,$ and $k$ be nonnegative real numbers. Prove that\n\\[(a^2+k+1)(b^2+k+1)(c^2+k+1) \\geq (k+2)^2(ab+bc+ca+k-1)\\]\nSetting $k=1$ gives the APMO 2004 problem.[/quote]\nhttp://artofproblemsolving.com/community/c6h453747p2550371\uff1a\n Let $a,b,c$ are real number and $k\\ge 2.$ Prove that \\[(a^2+k)(b^2+k)(c^2+k)\\ge (k+1)(a+b+c+k-2)^2.\\]", + "Solution_34": "[quote=ThAzN1]Let $a,b,c,$ and $k$ be nonnegative real numbers. Prove that\n\n\\[(a^2+k+1)(b^2+k+1)(c^2+k+1) \\geq (k+2)^2(ab+bc+ca+k-1)\\]\n\nSetting $k=1$ gives the APMO 2004 problem.[/quote]\n\n[img]http://s10.sinaimg.cn/middle/0018zOAxgy70jotSMpbb9&690[/img]", + "Solution_35": "Let $a,b,c,d$ are real number and $k\\ge 3.$ Prove or disprove \\[(a^2+k)(b^2+k)(c^2+k)(d^2+k)\\ge (k+1)^2(a+b+c+d+k-3)^2.\\]", + "Solution_36": "Let $a,b,c\\geq 0 .$ Prove that $$a^2+b^2+c^2+ab+bc+ca+2abc+1\\geq \\sqrt 3(a+b+c)$$\n\n$\\left( a+b+c\\right) \\left( \\left( 1+2abc+a^{2}+b^{2}+c^{2}\\right)-\\left(2bc+2ca+2ab\\right) \\right)$\n\n[url=https://artofproblemsolving.com/community/c4h2081311p14990870]Prove that[/url]\n\\[(a^2+k_1)(b^2+k_2)(c^2+k_3)\\geq \\frac{3k_1k_2k_3(a+b+c)^3}{4(k_1a+k_2b+k_3c)}\\]\\[(a^3+k_1)(b^3+k_2)(c^3+k_3)\\geq \\frac{3k_1k_2k_3(a+b+c)^3}{4(k_1+k_2+k_3)}\\]for any positiv real numbers $ a,b,c;k_1,k_2,k_3.$" +} +{ + "Tag": [ + "search", + "algebra", + "function", + "domain", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Why are the simle and the uniform convergence the same in the space of the polynom of degree equal or less than n ,where n is integer.", + "Solution_1": "The solution: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=471537135&t=114775[/url]", + "Solution_2": "Your statement in this thread is badly mangled. The bounded domain $ [0,1]$ is essential- if $ f_n$ converges to $ f$ uniformly on the line, $ f_n\\minus{}f$ is a constant for large enough $ n$.\r\n\r\nDo you understand the solutions in the linked thread?", + "Solution_3": "When i posted the problem i forgot to mention that the polynoms are conssidered in the cmpact interval .\r\nI tried to see in the beginning if any sequence in Rn[X] is equicontinuous on [0,1] to cnclude with ascoli theorem but i failed to do it and skipped the problem.\r\nWHY fn is equal to f when n is big enough.\r\nThks." +} +{ + "Tag": [ + "topology", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Suppose $ X$ is a complete metric space with metric $ d$ and $ T$ a self mapping of the space satisfying the following; $ d(x,y) n/2$, the term $ \\binom{n}{2i}\\binom{2i}{i}2^{n \\minus{} 2i}$ equals 0, because $ \\binom{n}{2i} \\equal{} 0$). Thus, the identity in question becomes\r\n\r\n$ \\sum_{i \\equal{} 0}^{n}\\binom{n}{2i}\\binom{2i}{i}2^{n \\minus{} 2i} \\equal{} \\binom{2n}{n}$.\r\n\r\nBut this was proven in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509115#509115]http://www.mathlinks.ro/Forum/viewtopic.php?t=87265 post #2[/url]. So Theorem 2 is proven. (There is also a more direct, counting proof of Theorem 2.)\r\n\r\nNow, in order to prove Theorem 1, we note that, for every integer k with $ 0\\leq k\\leq p \\minus{} 1$, we have\r\n\r\n[color=green][b](1)[/b][/color] $ \\binom{p \\minus{} 1}{k}\\equiv\\left( \\minus{} 1\\right)^{k}\\text{ mod }p$.\r\n\r\nThis is because\r\n\r\n$ \\binom{p \\minus{} 1}{k} \\equal{} \\frac {\\left(p \\minus{} 1\\right)\\cdot\\left(p \\minus{} 2\\right)\\cdot ...\\cdot\\left(p \\minus{} k\\right)}{1\\cdot 2\\cdot ...\\cdot k}\\equiv\\frac {\\left( \\minus{} 1\\right)\\cdot\\left( \\minus{} 2\\right)\\cdot ...\\cdot\\left( \\minus{} k\\right)}{1\\cdot 2\\cdot ...\\cdot k} \\equal{} \\left( \\minus{} 1\\right)^{k}\\text{ mod }p$.\r\n\r\nHereby, we legitimately worked modulo p, since the denominator of our fractions, $ 1\\cdot 2\\cdot ...\\cdot k$, is coprime to p (because k < p, and the number p is prime).\r\n\r\nAlso note that, for any integer k with $ \\frac {p \\minus{} 1}{2} < k\\leq p \\minus{} 1$, we have\r\n\r\n$ \\binom{2k}{k} \\equal{} \\frac {2k\\cdot\\left(2k \\minus{} 1\\right)\\cdot ...\\cdot\\left(k \\plus{} 1\\right)}{k\\cdot\\left(k \\minus{} 1\\right)\\cdot ...\\cdot 1}$;\r\n\r\nthe numerator of this fraction contains the factor p (since $ 2k > p \\minus{} 1$ and $ k \\plus{} 1\\leq p$) while the denominator doesn't (since k < p). Thus, the fraction is divisible by p, so we have $ \\binom{2k}{k}\\equiv 0\\text{ mod }p$ for every integer k with $ \\frac {p \\minus{} 1}{2} < k\\leq p \\minus{} 1$. Hence,\r\n\r\n$ \\sum_{k \\equal{} 0}^{\\frac {p \\minus{} 1}{2}}\\binom{2k}{k}\\left(\\frac12\\right)^{k} \\equal{} \\sum_{k \\equal{} 0}^{p \\minus{} 1}\\binom{2k}{k}\\left(\\frac12\\right)^{k} \\equal{} \\sum_{k \\equal{} 0}^{p \\minus{} 1}\\binom{2k}{k}2^{ \\minus{} k}\\equiv\\sum_{k \\equal{} 0}^{p \\minus{} 1}\\binom{2k}{k}2^{p \\minus{} 1 \\minus{} k}$ (since $ 2^{p \\minus{} 1}\\equiv 1\\text{ mod }p$)\r\n$ \\equal{} \\sum_{k \\equal{} 0}^{p \\minus{} 1}\\left( \\minus{} 1\\right)^{k}\\left( \\minus{} 1\\right)^{k}\\binom{2k}{k}2^{p \\minus{} 1 \\minus{} k}\\equiv\\sum_{k \\equal{} 0}^{p \\minus{} 1}\\left( \\minus{} 1\\right)^{k}\\binom{p \\minus{} 1}{k}\\binom{2k}{k}2^{p \\minus{} 1 \\minus{} k}$ (by [color=green][b](1)[/b][/color])\r\n$ \\equal{} T\\left(p \\minus{} 1\\right)\\text{ mod }p$ (by Theorem 2 for n = p - 1).\r\n\r\nNow, since p is odd, p - 1 is even, so that $ T\\left(p \\minus{} 1\\right) \\equal{} \\binom{p \\minus{} 1}{\\left(p \\minus{} 1\\right)/2}$. Also, [color=green][b](1)[/b][/color] yields $ \\binom{p \\minus{} 1}{\\left(p \\minus{} 1\\right)/2}\\equiv\\left( \\minus{} 1\\right)^{\\frac {p \\minus{} 1}{2}}\\text{ mod }p$. Thus, we get $ \\sum_{k \\equal{} 0}^{\\frac {p \\minus{} 1}{2}}\\binom{2k}{k}\\left(\\frac12\\right)^{k}\\equiv\\left( \\minus{} 1\\right)^{\\frac {p \\minus{} 1}{2}}\\text{ mod }p$, and Theorem 1 is proven.\r\n\r\n[b]EDIT:[/b] For another proof of Theorem 1, see Problem N.6 in Chapter 3 of\r\nG\u00e1bor J. Sz\u00e9kely (Editor), [i]Contests in Higher Mathematics[/i], New York 1996.\r\n\r\n Darij", + "Solution_2": "Before someone starts to think that this problem is THAT difficult, another solution:\r\n\r\n$(-1)^{\\frac{p-1}2}= (1-2)^{\\frac{p-1}2}= \\sum_{k=0}^{\\frac{p-1}2}{{\\frac{p-1}2}\\choose k}(-2)^{k}\\equiv \\sum_{k=0}^{\\frac{p-1}2}{{-\\frac{1}{2}}\\choose k}(-4)^{k}(\\frac{1}{2})^{k}= \\sum_{k=0}^{\\frac{p-1}2}{{2k}\\choose k}(\\frac{1}{2})^{k}$,\r\n\r\nsince a well-known identity states that ${{-\\frac{1}{2}}\\choose k}(-4)^{k}={{2k}\\choose k}$:\r\n\r\n${{-\\frac{1}{2}}\\choose k}(-4)^{k}= \\frac{(-\\frac{1}{2})(-\\frac{3}{2})\\cdots (-\\frac{2k-1}2)}{k!}(-4)^{k}= \\frac{1\\cdot 3\\cdots (2k-1)}{k!}2^{k}= \\frac{1\\cdot 3\\cdots (2k-1)}{k!}\\frac{2\\cdot 4\\cdots 2k}{1\\cdot 2\\cdots k}=\\frac{(2k)!}{(k!)^{2}}={{2k}\\choose k}$.\r\n\r\nPeter" +} +{ + "Tag": [ + "geometry", + "3D geometry", + "LaTeX", + "search", + "sphere" + ], + "Problem": "an ant/bug/snake/whatever you call :D , is in a corner of a room,\r\nhe has to reach diagonally opposite point \r\nhe can crawl on the walls only(no flying)\r\n\r\nfind the path he should take to reach that point in such a way that distance covered is minimum.\r\n\r\nI have been asked this question by my friend and now have the answer but i am not that much sure\r\n\r\nplz post your views", + "Solution_1": "::ahem that was the year's 1st prob of week at my insti\r\n\r\nthe ans is [code]$\\sqrt5$ a[/code]", + "Solution_2": "what ???????????!!!!!!!!!! :?: :huh:", + "Solution_3": "can you please elaborate??", + "Solution_4": "consider it in 2 dimensions\r\nflatten the cube\r\nif \"a\" is side of cube,\r\nshortest path will be (sq. root 5)a\r\n\r\ni don't kno exact latex of AOPS....... :rotfl:", + "Solution_5": "Why is this there in the general discussions? :huh: \r\n\r\nRefer the figure :) \r\n\r\n( i have used $ fa$ and $ (1 \\minus{} f)a$ instead of some $ x$ and $ (a \\minus{} x)$\r\nto lessen the computations :P )\r\n\r\nThe total distance covered is $ d \\equal{} \\sqrt {a^2 \\plus{} a^2f^2} \\plus{} \\sqrt {a^2 \\plus{} a^2(1 \\minus{} f)^2}$\r\n\r\nall you have got to do is minimise this expression for d. You'll find out that f=1/2 :)", + "Solution_6": "@ ritu\r\n\r\ndont use \"code\" da! just enclose the latex command within dollar signs! :D :lol:", + "Solution_7": "sakand---if u asked that question u are smartedr by a 1000 times than me.....i don't get a word-i mean number of this!!!! but it wasn't my question.", + "Solution_8": "thanks to all of you\r\n\r\nbtw,question was dealing with cubiodal roo(with different length and breadth :D )\r\ni understood,thnx again\r\n\r\n@crazy4math\r\n1.I am in 11th standard(or grade perhaps,i am not sure) and senior to you by at-least 3 standards :D)\r\n2.I am [b]skand[/b] not sakand :rotfl:", + "Solution_9": "Da Vince --i told a much simpler method :P \r\n\r\njust open the cube frm one edge and flatten it\r\nthen u can easily [b]see[/b] the shortest line :D", + "Solution_10": "sorry! I couldnt get you earlier :P :P :P", + "Solution_11": "SKAND----now how does that work cuz it is impossible to not have a vowel between s and k so i am right.....i hate being wrong and i can argue my point so just try me...and yes will go DOWN!!!!!!!!!!", + "Solution_12": "SKAND works,you may have argued about Skanda\r\n\r\nskand was the name of elder son of god shiva\r\nbtw,you can also search google about it :rotfl:", + "Solution_13": "[quote=\"rituraj007\"]c\ni don't kno exact latex of AOPS....... :rotfl:[/quote]\r\nthalai u r here froma century da and yet u dunno how to $ \\text{\\LaTeX}$ this in itself tells what we doa t AoPS :P :P :rotfl: :rotfl:", + "Solution_14": "wat was that languaage!", + "Solution_15": "AOPSian :rotfl: :rotfl:", + "Solution_16": "evn i dont know LATEX (correctly......)", + "Solution_17": "skand-skand was the name????of a god it was spelled that way. okay i lose. sorry, i seem to be doing everything wrong i really should leave the forum. :noo: :( :(", + "Solution_18": "the actual name was kartikeya(if you know)\r\n\r\nbtw,at this forum noone name me as Askand or sth like it\r\n\r\nmoreover i don't mind,my name is little weird and i have experience of people spelling incorrectly :rotfl:", + "Solution_19": "@crazy _ur going too much y r u makin fuss of all this \r\nwe all know that u don nthin wrong but still overblame urself.................", + "Solution_20": "hmm sound like me...... oh well\r\ni eel like posting math: wat's four thirds pi r squared the formula of???? that's the most newest thing i learned, yes i know u r not amazed.", + "Solution_21": "Volume of sphere per unit radius!! :rotfl: :rotfl: :rotfl:", + "Solution_22": "that was bahut easy.................", + "Solution_23": ":rotfl: :rotfl: my answer isn't the rite one!!!!!", + "Solution_24": "PER UNIT RADIUS :rotfl: :rotfl: :rotfl:", + "Solution_25": "it's the volume of a sphere. is that wat u said.", + "Solution_26": "vol of sphere is \r\n4/3 pi r cube :rotfl: :D", + "Solution_27": "that is the correct formula :arrow:", + "Solution_28": "yes certainly(though my memory is bit low)\r\ni can bet my last penny on this formula\r\n\r\nvol. of sphere is $ \\frac{4}{3}\\pi r^3$", + "Solution_29": ":rotfl: :rotfl: :rotfl:", + "Solution_30": "i'll bet my last \"paisa\" :P :oops:", + "Solution_31": ":rotfl: :rotfl: :rotfl: u dont hav to bet on that......................", + "Solution_32": "ha ha i meant cubed.................. :rotfl: :blush:", + "Solution_33": "hey wats up?\r\neverythin finee \r\n\r\nnow thats a spam?\r\nby me?", + "Solution_34": "wait what??? i don't get it?", + "Solution_35": "u will never get it :idea: :roll: :huh:", + "Solution_36": "i know i'm so stupid :( WAAH. boo hoo....throws a temper tantrum.", + "Solution_37": "sorry well let me not spam any more (except this one)", + "Solution_38": "more math? true of alse? the measure of an anhle formed by a chord and a tangent is equal to half the measure od the intercepted arc.", + "Solution_39": "false :rotfl: :rotfl: :rotfl:" +} +{ + "Tag": [ + "LaTeX", + "Gauss" + ], + "Problem": "Who made the formula $\\frac{(n)(n+1)}{2}$?\r\nI heard it was from a 4 year old.", + "Solution_1": "[quote=\"236factorial\"]Who made the formula $\\frac{(n)(n+1)}{2]}$? \nI heard it was from a 4 year old.[/quote]\r\n$\\frac{(n)(n+1)}{2}$\r\nYou got \"dangerous latex formula error\"\r\nGauss", + "Solution_2": "Umm... if his formula didn't appear, how can you know which formula he was referring to?", + "Solution_3": "[quote=\"JesusFreak197\"]Umm... if his formula didn't appear, how can you know which formula he was referring to?[/quote]\r\n\r\nYou can see the $\\LaTeX$ when you quote. :) By the way, Gauss was like 7 or 9, can't remember.", + "Solution_4": "For his life story go here: http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Gauss.html\r\n\r\nAnd yes he was 7, when he derived at that formula.", + "Solution_5": "well, its not[i] that[/i] hard to derive :D", + "Solution_6": "but it is way harder if no one has done it before because your not so sure your going in the right direction and if it was so easy how come no one before him derived it?", + "Solution_7": "[quote=\"JesusFreak197\"]Umm... if his formula didn't appear, how can you know which formula he was referring to?[/quote]\r\n\r\nIts in the title :P", + "Solution_8": "[quote=\"GoBraves\"][quote=\"JesusFreak197\"]Umm... if his formula didn't appear, how can you know which formula he was referring to?[/quote]\n\nIts in the title :P[/quote]\r\nI think a mod changed the title before his post :P", + "Solution_9": "[quote=\"noneoftheabove\"]but it is way harder if no one has done it before because your not so sure your going in the right direction and if it was so easy how come no one before him derived it?[/quote]\n\nWhen you're 7, you don't really know what's right or wrong yet. You assume everything right. But I haven't been 7 for 6 years. \n\n[quote=\"nat mc\"]\nI think a mod changed the title before his post :P[/quote]\r\n\r\nObviously, I don't think I ever wrote that :lol:", + "Solution_10": "The \"triangular\" numbers have been studied and known for a very long time. The school of Pythagoras took a particular interest in them (and the square numbers, and the pentagonal numbers, and so on) more than 2500 years ago.\r\n\r\nThe legend about the young Carl Gauss and his escape from his schoolmaster's tedious assignment is told [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Gauss.html]here[/url].", + "Solution_11": "[quote=\"Kent Merryfield\"]The \"triangular\" numbers have been studied and known for a very long time. The school of Pythagoras took a particular interest in them (and the square numbers, and the pentagonal numbers, and so on) more than 2500 years ago.\n\nThe legend about the young Carl Gauss and his escape from his schoolmaster's tedious assignment is told [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Gauss.html]here[/url].[/quote]\n\n Kent, when you posted that link did you fail to see myn?\n\n\n[quote=\"G-UNIT\"]For his life story go here: http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Gauss.html\n\nAnd yes he was 7, when he derived at that formula.[/quote]\r\n\r\n i guess no one reads my posts......... :noo: :read: :whistling: :agent:", + "Solution_12": "Gauss was 7? I find that amazing as I only recall being able to do fractions when I was 7.", + "Solution_13": "I'm sorry for not saying who fixed the latex and changed the title. I got a little distracted and hurried, and I forgot to note that. :oops_sign: \r\n\r\nBilly", + "Solution_14": "As young historians, be careful of the legends that you hear about mathematicians. They are sometimes either not true, or the details have been distorted. This is the case with many famous stories in mathematics.\r\n\r\nAccording to E.T. Bell, Gauss was 10 when he was given the problem, and it was actually a little bit harder than adding up the numbers from 1 to 100. [url=http://mathforum.org/social/articles/ross.html]See here[/url].\r\n\r\nBut E.T. Bell has also been accused of exaggerating the truth for a good story in some of his biographies so it\u2019s hard to know if his story is true either. The point, basically, (and this is true of history in general) is to only believe stories that come from source documents - like an actual letter by Gauss' teacher or someone in the class, or from someone that can cite source documents.\r\n\r\nBut as Kent mentioned, the formula for the sum of consecutive integers was known to the ancient Greeks, Chinese, and possibly others. The first to give a rigorous proof of it, though, may have been Pascal - as he was the father of mathematical induction.", + "Solution_15": "There's an interesting bit of minutia [url=http://www.answers.com/main/ntquery?method=4&dsid=2218&dekey=s04.2.90&gwp=8&curtab=2218_1&linktext=1575%20Mathematics]here[/url] about proving the sum of consecutive odd integers is $n^2$.", + "Solution_16": "It would be most wise to consider what Gauss has said. For no one knows better than the man himself. :lol:", + "Solution_17": "that's amazing how he found the formula at 10 years of age.", + "Solution_18": "[quote=\"math92\"]that's amazing how he found the formula at 7 years of age.[/quote]\r\n\r\nDid you not just read the post of gauss himself? :P", + "Solution_19": "I edited it, and no i didn't read his post. :P", + "Solution_20": "[quote=\"math92\"]I edited it, and no i didn't read his post. :P[/quote]\r\n\r\n well you still haven't understood what he was saying, gauss202 was trying to make tha point that these legends we base assumptions on, ( i.e gauss was 7 years or 10 years of age when he derived the formula) are not always accurate, unless we have some real documented proof from person(s) who witnessed the incident, but even these can alter the truth." +} +{ + "Tag": [ + "linear algebra", + "matrix", + "inequalities unsolved", + "inequalities" + ], + "Problem": "old problem,but good", + "Solution_1": "It seems that your problem disappeared. :(", + "Solution_2": "[u][b]Problem:[/b][/u] \\[\\begin{pmatrix}a_{1,1}\\ a_{1,2}\\ a_{1,3}\\ ... \\ a_{1,n}\\\\ a_{2,1}\\ a_{2,2}\\ a_{2,3}\\ ... \\ a_{2,n}\\\\ a_{3,1}\\ a_{3,2}\\ a_{3,3}\\ ... \\ a_{3,n}\\\\ \\\\ a_{k,1}\\ a_{k,2}\\ a_{k,3}\\ ... \\ a_{k,n}\\\\ \\\\ a_{n,1}\\ a_{n,2}\\ a_{n,3}\\ ... \\ a_{n,n}\\end{pmatrix}, \\{x_{i}\\}_{i=1}^{n}\\ s.t. \\ x_{i}\\in \\{-1,1\\}\\] We have that $\\sum_{k=1}^{n}\\left[\\sum_{i=1}^{n}\\left( a_{k,i}x_{i}\\right) \\right]\\leq M$\r\n[b][i]Prove[/i][/b] \\[\\sum_{j=1}^{n}a_{j,j}\\leq M\\]" +} +{ + "Tag": [ + "calculus", + "integration", + "function", + "Euler", + "limit", + "trigonometry", + "derivative" + ], + "Problem": "OK, this is an integral i like a lot... and to-date, i know of 7 ways to solve it,\r\nthey are, using:\r\n\r\n(1) partial fraction\r\n(2) trigonometric substitution\r\n(3) residue calculus (cauchy)\r\n(4) beta function (euler)\r\n(5) fourier transform\r\n(6) DuIs (leibnitz)\r\n(7) IuIs (newton)\r\n\r\nif you know of any other method, let me know, OK? \r\n\r\nbelow, i will show only (4), but not completely, 'cause i'm too lazy to type! :rotfl:\r\n\r\nPS: it, in fact, also links to (3).\r\n\r\n$\\mathbb{B}\\;(m,n)\\;=\\;\\int_{0}^{1}\\;x^{m-1}\\;(1-x)^{n-1}\\;dx\\;\\quad\\; (\\;where\\;\\;m,n\\;>\\;0\\; )$\r\n\r\nput $x=\\frac{1}{1+y}$ and it gives\r\n\r\n$\\mathbb{B}\\;(m,1-m)\\;=\\;\\int_{0}^{\\infty}\\;\\frac{y^{m-1}}{1+y}\\;dy\\;\\quad\\; (\\;where\\;\\;0\\;<\\;m\\;<\\;1\\; )$\r\n\r\nput $y=\\frac{1}{u}$, and we get\r\n\r\n$\\int_{0}^{\\infty}\\;\\frac{y^{m-1}}{1+y}\\;dy\\;=\\;\\int_{0}^{\\infty}\\;\\frac{u^{-m}}{1+u}\\;du\\;\\;\\;\\quad ...\\;\\;\\;...\\;\\;\\;(1)$\r\n\r\n\r\nthe associated complex function $f(z)$ has $z^{-m}$ in the numerator,\r\nwhich is a multivalued function that needs a branch cut\r\nspecified, and for convenience, we will use the primary branch\r\n$0\\;\\leq\\;\\angle z\\;<\\;2\\pi$.\r\n\r\nusing the C-shaped contour (indented at $z=0$), we get\r\n\r\n$(1-e^{-\\imath 2m\\pi})\\;\\lim_{\\epsilon\\rightarrow 0,R\\rightarrow\\infty}\\;\\int_{\\epsilon}^{R}\\;\\frac{u^{-m}}{1+u}\\;du\\;=\\;2\\pi\\imath\\;e^{-\\imath m\\pi}\\;$\r\n\r\n\r\nwhich when simplified yields,\r\n\r\n$\\int_{0}^{\\infty}\\;\\frac{u^{-m}}{1+u}\\;du\\;=\\;\\frac{\\pi}{\\sin\\; (m\\pi)}$\r\n\r\ntherefore, we get, from $(1)$,\r\n\r\n$\\int_{0}^{\\infty}\\;\\frac{y^{m-1}}{1+y}\\;dy\\;=\\;\\frac{\\pi}{\\sin\\; (m\\pi)}$\r\n\r\nnow put, $y\\;=\\;v^{q}$, and $m\\;=\\;\\frac{p}{q}\\;$, and you will get\r\n\r\n$\\int_{0}^{\\infty}\\;\\frac{v^{p-1}}{1+v^{q}}\\;dv\\;=\\;\\frac{\\pi}{q}\\cdot\\;\\frac{1}{\\sin\\; ( \\frac{p\\;\\pi}{q}) \\;}\\;\\;\\; ( \\;where\\;\\; p,q\\;\\in\\;\\mathbb{R}^{+},\\;\\;p\\;<\\;q\\; )$\r\n\r\ni.e.\r\n\r\n$\\int_{0}^{\\infty}\\;\\frac{x^{p}}{1+x^{q}}\\;dx\\;=\\;\\frac{\\pi}{q}\\cdot\\;\\frac{1}{\\sin\\; ( \\frac{(p+1)\\;\\pi}{q}) \\;}$\r\n\r\nwhich means, also\r\n\r\n$\\int_{0}^{\\infty}\\;\\frac{x^{2p}}{1+x^{2q}}\\;dx\\;=\\;\\frac{\\pi}{2q}\\cdot\\;\\frac{1}{\\sin\\; ( \\frac{(2p+1)\\;\\pi}{2q}) \\;}$\r\n\r\n\r\n :coolspeak:", + "Solution_1": "${x=\\left(e^{y}-1\\right)^{1/b}}$\r\n${dx=\\frac{e^{y}\\left(-1+e^{y}\\right)^{\\frac{1}{b}-1}}{b}dy}$\r\n${\\gamma =\\frac{a+1}{b}-1}$\r\n\r\n${\\frac{x^{a}}{x^{b}+1}dx=\\frac{1}{b}\\left(-1+e^{y}\\right)^{\\frac{a+1}{b}-1}dy=\\frac{1}{b}\\left(-1+e^{y}\\right)^{\\gamma }dy}$\r\n\r\n${\\int_{0}^{\\infty }\\frac{x^{a}}{x^{b}+1}\\, dx=\\frac{1}{b}\\int_{0}^{\\infty }\\left(e^{y}-1\\right)^{\\gamma }\\, dy}$\r\n\r\n${\\{b>1,a>1,-1<\\gamma <0\\}}$\r\n\r\nA really waste of time\r\n\r\nhumpf\r\n\r\nhauhau, but is beautiful", + "Solution_2": "Dear misan, would you please explain, what are \n(6) DuIs (leibnitz) and (7) IuIs (newton) ?", + "Solution_3": "[quote=\"Fedor Petrov\"]Dear misan, would you please explain, what are \n(6) DuIs (leibnitz) and (7) IuIs (newton) ?[/quote]\nI suppose it means Derivation and Integration under the integral sign respectively.", + "Solution_4": "Thanks, but how do we use it here?", + "Solution_5": "One can see my reference [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=342117&]here[/url] for too methods of the special case $\\int_{0}^{+\\infty}\\frac{x^{p-1}}{1+x}\\,dx$. The second method is via differentiation if I remember well..." +} +{ + "Tag": [], + "Problem": "[color=darkblue]A rabbit is persecuted by a dog. The rabbit takes an initial advantage of 50 of its jumps to the dog. The rabbit gives 5 jumps whereas the dog gives 2, but the dog in 3 jumps advances so much as the rabbit in 8 jumps. How many jumps must give the dog to reach the rabbit?[/color]\r\n\r\n[hide=\"Condition\"]\n[color=darkred]It is only allowed to use a single variable to solve the problem[/color][/hide]", + "Solution_1": "Do you mean that the rabbit does 5 jumps for every 2 the dog does, but the [i]rabbit[/i] in 3 jumps advances as much as the dog in 8 jumps?\r\nAlso, what do you mean by an initial advantage of 50 jumps? You mean it starts 50 of its jumps behind where the dog is at the beginning?", + "Solution_2": "[quote=\"mets501\"]Do you mean that the rabbit does 5 jumps for every 2 the dog does, but the [i]rabbit[/i] in 3 jumps advances as much as the dog in 8 jumps?[/quote]\r\nThen the dog will never be able to catch up :( I think it's the other way.", + "Solution_3": "Typo :rotfl: \r\n\r\nNow is correct.", + "Solution_4": ":D funny, single variables only\r\n\r\n[hide=\"answer\"]\n\nMultiply 8*2 and 5*3 and the dog can go 16 jumps while the rabbit goes 15. He has to catch up 50 yards, so it will take him 48*16+1. It is 48 instead of 50 because after 48 sets the dog will leap and catch the poor little rabbit :) go dogs.\n\nI am temporarily without a calculator, so I have to do it by myself :( 769 is the answer.[/hide]", + "Solution_5": "[color=darkred]Sorry, is not correct.[/color]", + "Solution_6": "[quote=\"MathTeX\"][color=darkblue]A rabbit is persecuted by a dog. The rabbit takes an initial advantage of 50 of its jumps to the dog. The rabbit gives 5 jumps whereas the dog gives 2, but the dog in 3 jumps advances so much as the rabbit in 8 jumps. How many jumps must give the dog to reach the rabbit?[/color][/quote]\r\n[hide=\"Solution\"]Each jump the dog takes is 8 feet (for simplicity), and each jump the rabbit takes is 3 feet. Each time, the dog jumps twice (16 ft), and the rabbit jumps 5 times (15 ft). Every 2 jumps, the dog gains 1 ft on the rabbit. In the beginning, the rabbit is 50 jumps (150 ft) ahead of the dog. The dog will need to make $\\boxed{300\\text{ jumps}}$ to reach the rabbit.[/hide]\r\n@ MathTex - Please don't use color when writing problems." +} +{ + "Tag": [ + "LaTeX" + ], + "Problem": "how can you download Latex? I tried to download it from the links given in AoPS wiki, but failed.", + "Solution_1": "Which version/link did you try and how did it fail?", + "Solution_2": "Hi Attila,\r\n\r\nalternatively have a look at [url=http://texblog.net/latex-link-archive/distribution-editor-viewer/]Distributions, Editors, Viewers for LaTeX[/url], choose your operating system, you will find several alternative programs for each purpose. I'm preferring Kile with TeX Live 2008 on Ubuntu Linux, with Windows I'm using TeXnicCenter with TeX Live 2008 but MiKTeX is recommendable too. You will find downloads if you follow my link above.\r\n\r\nStefan\r\n\r\n--\r\n[url=http://texblog.net/]TeXblog.net[/url]" +} +{ + "Tag": [ + "probability" + ], + "Problem": "how would u do this problem?\r\n\r\nin a bag, there are 3 red marbles and B blue marbles. two marbles are randomly selected from the bag without replacement. the probability that the two marbles are the same color is 0.5. what calculate the sum of all the possible values for B.", + "Solution_1": "[hide]$\\frac{3\\cdot2+B\\cdot(B-1)}{(3+B)(2+B)}=.5$\n$6+B^{2}-B=.5B^{2}+2.5B+3$\n$.5B^{2}-3.5B+3=0$\n$B^{2}-7B+3=0$\n$(B-1)(B-6)=0$\n$B=1,6$[/hide]", + "Solution_2": "[hide]The probability of getting 2 red is $\\frac{3}{B+3}\\times \\frac{2}{2+B}$ Similarily, the probability of getting 2 blue is $\\frac{B}{3+B}\\times \\frac{B-1}{2+B}$ Then the total probability is $\\frac{6+B^{2}-B}{(3+B)(2+B)}=\\frac{1}{2}$ Then we have that $2B^{2}-2B+12=B^{2}+5B+6$, or $B^{2}-7B+6=0=(B-6)(B-1)$ Therefore, B can be 1 or 6.[/hide]", + "Solution_3": "So the sum of $B$ would be $7$. :wink:", + "Solution_4": "[quote=\"buzzer11\"]So the sum of $B$ would be $7$. :wink:[/quote]\r\nWhy? I don't understand. Please show me more concrete. Thanks a lot.", + "Solution_5": "[quote=\"mr. math\"]calculate the sum of all the possible values for B.[/quote]\r\n\r\nI should have said the sum of all possible values for $B$.\r\n\r\n :: :) ::" +} +{ + "Tag": [ + "factorial", + "search", + "number theory unsolved", + "number theory" + ], + "Problem": "Solve the equation in integers : $ \\ x^2 \\plus{} 2 \\equal{} y^3$\r\n\r\nPlease help!!!", + "Solution_1": "Factor it in $ \\mathbb Z[\\sqrt { \\minus{} 2}]$ and use that it is a factorial domain. Or simply use search function.\r\nFor the future you may want to use titles that tell something about the problem. \"Equation\" doesn't.", + "Solution_2": "I am very sorry about the title. I found solutions 5,3 and -5,3. Are there then any more?", + "Solution_3": "These are the only solutions, yes." +} +{ + "Tag": [ + "inequalities", + "trigonometry", + "limit", + "integration", + "inequalities proposed" + ], + "Problem": "Prove that there exists $ k > 0$ such that $ |{\\sum\\limits_{i \\equal{} 1}^n {\\frac{{\\sin i}}{i}} }| < k$ for every $ n\\in \\mathbb N$", + "Solution_1": "No solution for this :huh:", + "Solution_2": "[quote=\"N.N.Trung\"]Prove that there exists $ k > 0$ such that $ |{\\sum\\limits_{i = 1}^n {\\frac {{\\sin i}}{i}} }| < k$ for every $ n\\in \\mathbb N$[/quote]\r\nBecause, ${ \\lim_{n\\rightarrow+\\infty}\\sum_{i=1}^n{\\frac {{\\sin i}}{i}} }=\\frac{\\pi-1}{2}.$ :wink:", + "Solution_3": "[quote=\"arqady\"][quote=\"N.N.Trung\"]Prove that there exists $ k > 0$ such that $ |{\\sum\\limits_{i = 1}^n {\\frac {{\\sin i}}{i}} }| < k$ for every $ n\\in \\mathbb N$[/quote]\nBecause, ${ \\lim_{n\\rightarrow + \\infty}\\sum_{i = 1}^n{\\frac {{\\sin i}}{i}} } = \\frac {\\pi - 1}{2}.$ :wink:[/quote]\r\n\r\nplease tell me why?", + "Solution_4": "[quote=\"thuyanh158\"][quote=\"arqady\"][quote=\"N.N.Trung\"]Prove that there exists $ k > 0$ such that $ |{\\sum\\limits_{i = 1}^n {\\frac {{\\sin i}}{i}} }| < k$ for every $ n\\in \\mathbb N$[/quote]\nBecause, ${ \\lim_{n\\rightarrow + \\infty}\\sum_{i = 1}^n{\\frac {{\\sin i}}{i}} } = \\frac {\\pi - 1}{2}.$ :wink:[/quote]\n\nplease tell me why?[/quote]\r\nBecause $ \\sum_{i = 1}^n\\frac {\\sin ix}{i} = \\int_0^x\\sum_{i = 1}^n\\cos{it}dt = \\int_0^x\\sum_{i = 1}^n\\frac {2\\cos{it}\\sin{\\frac {t}{2}}}{2\\sin{\\frac {t}{2}}}dt =$\r\n$ = \\int_0^x\\sum_{i = 1}^n\\frac {\\sin{\\frac {(2i + 1)t}{2}} - \\sin{\\frac {(2i - 1)t}{2}}}{2\\sin{\\frac {t}{2}}}dt = \\int_0^x\\frac {\\sin{\\frac {(2n + 1)t}{2}} - \\sin{\\frac {t}{2}}}{2\\sin{\\frac {t}{2}}}dt =$\r\n$ = - \\frac {x}{2} + \\int_0^x\\left(\\frac {1}{2\\sin\\frac {t}{2}} - \\frac {1}{t}\\right)\\sin{\\frac {(2n + 1)t}{2}dt + \\int_0^x\\frac {\\sin{\\frac {(2n + 1)t}{2}}}{t}dt =}$\r\n$ = - \\frac {x}{2} + \\int_0^x\\left(\\frac {1}{2\\sin\\frac {t}{2}} - \\frac {1}{t}\\right)\\sin{\\frac {(2n + 1)t}{2}dt + \\int_0^{\\frac {(2n + 1)x}{2}}\\frac {\\sin t}{t}dt.}$\r\nThus, $ \\lim_{n\\rightarrow + \\infty}\\sum_{i = 1}^n\\frac {\\sin ix}{i} = - \\frac {x}{2} + 0 + \\frac {\\pi}{2} = \\frac {\\pi - x}{2}.$ :wink:" +} +{ + "Tag": [ + "number theory unsolved", + "number theory" + ], + "Problem": "Find all the positive integers a,b,c,d such that\r\n$ 1 -epi*i ", + "Solution_4": "[quote=\"tom11784\"]after two years of not helping much, maybe the third will be better if we get enough that they can handle a few more student coaches than this past year -epi*i [/quote]\r\n\r\nIt's comments like this that really remind me of why this forum was such a good idea :-D. Thanks Alison!", + "Solution_5": "You're welcome, although the AoPS folks did most of the work.", + "Solution_6": "they would not have if you hadn't asked them" +} +{ + "Tag": [], + "Problem": "The numbers $ x^3$ and $ y^2$ are inversely proportional. If $ x\\equal{}2$ when $ y\\equal{}15$, what is $ y$ when $ x\\equal{}1$?", + "Solution_1": "If two numbers $ x$ and $ y$ are inversely proportional, this means the product $ xy$ is always the same. So if $ x$ increases, $ y$ decreases.\r\n\r\nIf $ x \\equal{} 2$ when $ y \\equal{} 15$, the constant product is $ 2^3 \\cdot 15^2 \\equal{} 1800$. So when $ x \\equal{} 1$, $ y$ is $ \\sqrt {1800}$ or $ \\boxed{30\\sqrt{2}}$.\r\n\r\nEDIT: Oops, thanks for pointing that out. Fixed. :)", + "Solution_2": "[quote=\"isabella2296\"]If two numbers $ x$ and $ y$ are inversely proportional, this means the product $ xy$ is always the same. So if $ x$ increases, $ y$ decreases.\n\nIf $ x \\equal{} 2$ when $ y \\equal{} 15$, the constant product is $ 2^3 \\cdot 15^2 \\equal{} 1800$. So when $ x \\equal{} 1$, $ y$ is $ \\sqrt {8100}$ or $ \\boxed{90}$.[/quote]\r\n\r\nLooks like a transposition occured. That should read:\r\n\r\n$ y$ is $ \\sqrt {1800}$ or $ \\boxed{30\\sqrt{2}}$." +} +{ + "Tag": [ + "calculus", + "derivative", + "vector", + "function", + "calculus computations" + ], + "Problem": "Let $\\alpha: I \\to \\mathbf{R^{3}}$ be a diferentiable parametrized curve. Show that $\\alpha(t) \\bot \\alpha^{\\prime}(t)$ for all $t \\in I$ iff, $|\\alpha(t)|=a$ for all $t \\in I$ and some positive real $a$.\r\n\r\nMaybe it's too simple but, i got stuck because i got a quocient on the derivative of $|\\alpha(t)|$ wich not exists when $\\alpha(t)=(0,0,0)$. Thanks for any help", + "Solution_1": "Lesson number 1: squares of magnitudes are easier to deal with than magnitudes themselves, and given the chance, [b]always[/b] write the square of a magnitude as a dot product: $|v|^{2}=v\\cdot v.$\r\n\r\nLesson number 2: all of those vector products satisfy easily predictable product rules.\r\n\r\nGiven that:\r\n\r\n$\\frac{d}{dt}|\\alpha|^{2}=\\frac{d}{dt}(\\alpha\\cdot\\alpha)=\\alpha'\\cdot\\alpha+\\alpha\\cdot\\alpha'$\r\n\r\n$=2\\alpha\\cdot\\alpha'$ since the dot product is commutative.\r\n\r\nSo from that, $|\\alpha|$ is constant iff $\\frac{d}{dt}|\\alpha|^{2}=0$ iff $\\alpha\\cdot\\alpha'=0,$ which is true iff $\\alpha\\perp\\alpha'.$\r\n\r\nEssentially the same argument shows that a particle in motion has constant speed if and only if its acceleration is always perpendicular to its velocity.", + "Solution_2": "Very thanks Mr. Kent, much easier and elegant this way. Good lessons. and nice aplication. In my solution i set $f(t)=|\\alpha(t)|$ so $f'(t)=\\frac{\\alpha'\\cdot \\alpha }{|\\alpha(t)|}$. If $\\alpha(t)\\neq (0,0,0)$ then $f'(t)=0$ and $|\\alpha(t)|$ is equal to some positive real. But if $\\alpha(t) = (0,0,0)$ then $f'$ is not defined. So i know there must be some way to go from here... i'm curious...\r\nAnyway thanks again :)", + "Solution_3": "Notice that the original statement ruled out what is happening at the origin.\r\n\r\nIf $\\alpha=0$ then the meaning of $\\alpha\\perp\\alpha'$ is unclear although the dot product is still zero. But notice the statement $|\\alpha(t)|=a$ for some [b]positive[/b] real $a.$ That takes the origin out of the picture.\r\n\r\nThe function $|\\alpha(t)|$ is not differentiable when $\\alpha(t)=0.$ The function $|\\alpha(t)|^{2}$ is differentiable and has derivative zero at such a point.", + "Solution_4": ":oops: Oh that's right! $|\\alpha|=0$ iff $\\alpha=0$. That analisys was unnecessaire, and the square magnitude is really better giving the proof to $\\alpha=0$ too. Much thanks Mr. Kent, now i can sleep in peace." +} +{ + "Tag": [ + "Gauss" + ], + "Problem": "Fie A o matrice patratica cu elemente rationale astfel incat pentru un $ k>1$ avem $ A^k=I$. Atunci polionomul caracteristic al lui A are coeficienti intregi. Are cineva o solutie care nu foloseste ireductibilitatea polinoamelor ciclotomice?", + "Solution_1": "Imi inchipui ca ar trebui sa fie destul de nasol fara asta :?\r\n\r\nPana una-alta :),\r\n\r\nPolinomul caracteristic e rational si divide $x^k-1=\\prod_{d|k}\\phi_d(x)$. Cum $\\phi_d$ sunt ireductibile, inseamna ca polinomul caracteristic al lui $A$ e asociat cu un produs de polinoame ciclotomice, si cum astea sunt monice si la fel e si polinomul caracteristic al lui $A$, inseamna ca acesta e chiar produsul pol. Ciclotomice care il divid, deci are coeficienti intregi. \r\n\r\nStiu ca nu voiai solutia asta, dar presupun ca fara ireductibilitatea polinoamelor ciclotomice, oricum cred (:?) ca ar trebui sa incropim ceva care seamana cu demonstratia ireductibilitatii, deci tot acolo am ajunge.", + "Solution_2": "Eu stiu ca demonstratia ireductibilitatii polinoamelor ciclotomice e o treaba cu adevarat grea in draci. Desigur, am o demonstratie in mai multi pasi folosind teorema lui Dirichlet, dar ma gandesc ca ar trebui sa fie si o solutie normala de au dat-o astia la oral (am uitat sa spun ca la o proba orala cam trebuie sa stii sa demonstrezi tot ce folosesti-or demonstratia ireductibilitatii polinoamelor ciclotomice nu e tocmai un lucru placut).", + "Solution_3": "Bai nu imi vine sa cred ca nu am vazut asta pana acum... :?\r\n\r\nCe avem noi de demonstrat e ca un polinom rational monic care divide $x^k-1$ are, de fapt, coeficienti intregi, nu? Pai sa presupunem ca $P(x)\\in\\mathbb Q$ e monic si divide $x^k-1$. Atunci, radacinile lui sunt intregi algebrici, deci, fiind monic si scris ca $(x-x_1)\\ldots (x-x_t)$, inseamna ca toti coeficientii lui sunt intregi algebrici, pentru ca sunt sume de produse de intregi algebrici $x_i$. De asemenea, stim ca toti coeficientii lui sunt rationali, si cum singurii intregi algebrici rationali sunt intregii...\r\n\r\nE ceva gresit pe aici? Parca e prea usor sa fie adevarat :).", + "Solution_4": "Superb, este super fain, grobber. Si era atat de natural, fir-ar al naibii... Mersi, e o adevarata perla.", + "Solution_5": "Pai sa stii ca nici mie nu-mi venea. Nu stiu ce naiba a avut problema asta :). Citeam o chestie in Teoria Numerelor de Borevici si Safarevici, o lema, si mi-am dat seama ca generaliza chestia pe care o avem noi aici.", + "Solution_6": "[quote=\"grobber\"]Ce avem noi de demonstrat e ca un polinom rational monic care divide $x^{k}-1$ are, de fapt, coeficienti intregi, nu?[/quote]\r\n\r\nDaca scriem $x^{k}-1 = P(x) Q(x)$, unde $P$ e polinom rational monic, folosind algoritmul de \"long division\" deducem c\u0103 \u015fi $Q$ e polinom rational monic. Lema lui Gauss ne spune ca exista $a,b \\in \\mathbb Q$ astfel incat $a P(x), \\, b Q(x) \\in \\mathbb Z[x]$ si $x^{k}-1 = (aP(x)) (bQ(x))$. Coeficientul dominant al lui $a P(x)$ (care trebuie sa fie intreg) este $a$, deci $a \\in \\mathbb Z$. Asemanator, $b \\in \\mathbb Z$. Cum $ab = 1$, rezulta ca $a = b = \\pm 1$, deci $P,Q \\in \\mathbb Z[x]$.\r\n\r\nE ceva gre\u015fit \u00een chestia asta? :maybe:\r\nChiar daca e buna rezolvarea, tot are unele dezavantaje :(" +} +{ + "Tag": [ + "algebra", + "polynomial", + "algebra proposed" + ], + "Problem": "Answer the following questions. You may use the fact that $ \\sqrt{2},\\ \\sqrt{3},\\ \\sqrt{6}$ are irrational without proof.\r\n\r\n(1) If $ p\\plus{}q\\sqrt{2}\\plus{}r\\sqrt{3}\\equal{}0$ for rational numbers $ p,\\ q,\\ r$, then prove that $ p\\equal{}q\\equal{}r\\equal{}0$.\r\n\r\n(2) Let $ f(x)\\equal{}x^2\\plus{}ax\\plus{}b$ be a polynomial with real coefficient. Prove that at least of $ f(1),\\ f(1\\plus{}\\sqrt{2}),\\ f(\\sqrt{3})$ is irrational number.", + "Solution_1": "(1) WLOG $ p,\\ q,\\ r$ are integers by taking common denominator.\r\n$ q\\sqrt {2} \\plus{} r\\sqrt {3} \\equal{} \\minus{} p$ \r\nTaking squares on both sides: \r\n$ (q\\sqrt {2} \\plus{} r\\sqrt {3})^2 \\equal{} p^2$\r\n$ 2q^2 \\plus{} 3r^2 \\plus{} 2qr \\sqrt {6} \\equal{} p^2$ \r\n$ \\sqrt {6} \\equal{} \\frac {p^2 \\minus{} 2q^2 \\minus{} 3r^2}{2qr}$ but $ \\sqrt {6}$ is irrational.\r\nSimilarly for $ qr \\equal{} 0$.", + "Solution_2": "[quote=\"mszew\"](1) WLOG $ p,\\ q,\\ r$ are integers by taking common denominator.\n$ q\\sqrt {2} \\plus{} r\\sqrt {3} \\equal{} \\minus{} p$ \nTaking squares on both sides: \n$ (q\\sqrt {2} \\plus{} r\\sqrt {3})^2 \\equal{} p^2$\n$ 2q^2 \\plus{} 3r^2 \\plus{} 2qr \\sqrt {6} \\equal{} p^2$ \n$ \\sqrt {6} \\equal{} \\frac {p^2 \\minus{} 2q^2 \\minus{} 3r^2}{2qr}$ but $ \\sqrt {6}$ is irrational.\nSimilarly for $ qr \\equal{} 0$.[/quote]\r\n\r\nYour solution is logically wrong.", + "Solution_3": "Assume p!=0.\r\n\r\nq*sqrt2+rsqrt3=-p\r\nq and r cannot both be 0 (or p=0, which is against the assumption), so if q or r are zero, q*sqrt2= -p or r*sqrt3= -p which are contradictory, since the left sides are irrational, and the right side is rational, thus q and r cannot be 0.\r\n\r\nSince it equals -p, q*sqrt2+r*sqrt3 must be rational.\r\nDivide by q, so sqrt2+r*sqrt3/q must be r/q multiplied by some rational number k.\r\n2+r*sqrt6/q=rk*sqrt2/q\r\n2=rk*sqrt2/q-r*sqrt6/q\r\n2=r/q(ksqrt2-sqrt6)\r\n2q/r=sqrt2(k-sqrt3)\r\nLeft side is rational, right side is irrational, thus q*sqrt2+r*sqrt3 cannot be rational, thus either p=0, or the equation has no solutions.\r\n\r\nAssume p=0;\r\nq*sqrt2+r*sqrt3=0\r\n\r\nEITHER q!=0\r\nq(sqrt2+r*sqrt3/q)=0\r\n\r\nsqrt2+r*sqrt3/q=0\r\nr*sqrt3= -q*sqrt2\r\n-r/q=sqrt2/sqrt3\r\n\r\n-r/q is rational, sqrt2/sqrt3 is not. This cannot be, so q=0 or the solution has no equations.\r\nIf p=0 and q=0,\r\nrsqrt3=0\r\nr=0\r\nGiving the desired equalities, p=q=r=0.", + "Solution_4": "[quote=\"EastyMoryan\"]Assume $ p\\neq 0$.\n\n$ q\\sqrt {2} \\plus{} r\\sqrt {3} \\equal{} \\minus{} p$\n$ q$ and $ r$ cannot both be 0 (or $ p \\equal{} 0$, which is against the assumption), so if $ q$ or $ r$ are zero, $ q\\sqrt2 \\equal{} \\minus{} p$ or $ r\\sqrt3 \\equal{} \\minus{} p$ which are contradictory, since the left sides are irrational, and the right side is rational, thus $ q$ and $ r$ cannot be 0.\n\n\nSince it equals $ \\minus{} p, q\\sqrt2 \\plus{} r\\sqrt3$ must be rational.\nMultiplying a rational by a rational gives a rational, so multiply by $ \\sqrt {2}.$\n$ 2q \\plus{} r\\sqrt {6}$ must be some rational number $ k$.\n$ 2q \\plus{} r\\sqrt {6} \\equal{} k$\n$ r\\sqrt6 \\equal{} k \\minus{}2q$ (Difference of two rationals is rational)\n$ \\sqrt {6} \\equal{} (k \\minus{} 2q)/r$ \nLeft side is irrational, right side is rational, thus $ q\\sqrt {2} \\plus{} r\\sqrt {3}$ cannot be rational, thus either $ p \\equal{} 0$, or the equation has no solutions.\n\nAssume $ p \\equal{} 0$;\n$ q\\sqrt {2} \\plus{} r\\sqrt {3} \\equal{} 0$\n\nEITHER $ q \\equal{} 0\\ \\ r\\sqrt3 \\equal{} 0\\ r \\equal{} 0$\n\nOR $ q(\\sqrt {2} \\plus{} r\\sqrt {3/q}) \\equal{} 0$\n\n$ \\sqrt {2} \\plus{} r\\sqrt {3/q} \\equal{} 0\\ r\\sqrt {3} \\equal{} \\minus{} q\\sqrt {2}\\ \\minus{} r/q \\equal{} \\sqrt {2}/\\sqrt {3}$\n\n$ \\minus{} r/q$ is rational, $ \\sqrt {2}\\ \\sqrt {3}$ is not. This cannot be, so $ p \\equal{} q \\equal{} r \\equal{} 0$.[/quote]", + "Solution_5": "Edited it ^_^\r\n\r\n[quote]Multiplying a rational by a rational gives a rational, so multiply by sqrt2[/quote]Doesn't work.}", + "Solution_6": "[quote=\"kunny\"][quote=\"mszew\"](1) WLOG $ p,\\ q,\\ r$ are integers by taking common denominator.\n$ q\\sqrt {2} \\plus{} r\\sqrt {3} \\equal{} \\minus{} p$ \nTaking squares on both sides: \n$ (q\\sqrt {2} \\plus{} r\\sqrt {3})^2 \\equal{} p^2$\n$ 2q^2 \\plus{} 3r^2 \\plus{} 2qr \\sqrt {6} \\equal{} p^2$ \n$ \\sqrt {6} \\equal{} \\frac {p^2 \\minus{} 2q^2 \\minus{} 3r^2}{2qr}$ but $ \\sqrt {6}$ is irrational.\nSimilarly for $ qr \\equal{} 0$.[/quote]\n\nYour solution is logically wrong.[/quote]\r\n \r\n(1)The solution of mszew is not so bad. :lol: . May be it needs a few details.\r\n$ qr\\neq 0$ leeds to the contradiction already proved .\r\n$ q\\neq 0,\\ r \\equal{} 0$ leeds to the contradiction $ \\sqrt 2 \\equal{} \\minus{} \\frac pq$\r\n$ q \\equal{} 0,\\ r\\neq 0$ leeds to the contradiction $ \\sqrt 3 \\equal{} \\minus{} \\frac pr$\r\nTherefore we have to have $ q \\equal{} r \\equal{} 0$, so $ p \\equal{} 0$ \r\n............\r\n(2) Let's suppose that $ f(1)\\equal{}p,\\ f(1\\plus{}\\sqrt{2})\\equal{}q,\\ f(\\sqrt{3})\\equal{}r$ are rational.\r\n$ (I)1\\plus{}a\\plus{}b \\equal{} p, \\ (II)3 \\plus{} 2\\sqrt 2 \\plus{} a(1\\plus{}\\sqrt 2) \\plus{} b\\equal{}q ,\\ (III)\\ 3 \\plus{} a\\sqrt 3 \\plus{} b \\equal{} r$\r\n$ (II)\\minus{}(I) \\Longrightarrow a \\equal{} U_1\\sqrt 2 \\plus{} U_2 \\ ;\\ (II)\\minus{}(I) \\Longrightarrow a \\equal{} V_1\\sqrt 3 \\plus{} V_2$ , where $ U_1,U_2,V_1,V_2$ are rational\r\nTherefore $ U_1\\sqrt 2 \\plus{} U_2 \\equal{} V_1\\sqrt 3 \\plus{} V_2$, and because of the first problem we obtain $ U_1\\equal{}V_1\\equal{}0$, so $ a$ is rational.\r\nBut $ b\\equal{}p\\minus{}1\\minus{}a$, so $ b$ is rational, and from (III) we obtain that $ a\\equal{}0$ . We replace in $ (II)$ and that leeds to the contradiction $ \\sqrt 2 \\equal{}\\frac{q\\minus{}b\\minus{}3}2$\r\nFinally at least of $ f(1),\\ f(1\\plus{}\\sqrt{2}),\\ f(\\sqrt{3})$ is irrational .\r\n\r\n :cool:" +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "Let $x,y,z>0$.Prove the inequality:\r\n\r\n $\\frac{xy}{x+y}+\\frac{yz}{y+z}+\\frac{zx}{z+x}\\geq\\frac{3xyz}{xy+yz+zx}+\\frac{xy+yz+zx}{2(x+y+z)}$ ;)", + "Solution_1": "Using substitution $a=1/x,\\,b=1/y,\\,c=1/z$ we obtain the equivalent inequality\r\n\\[\r\n\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}\\geq \\frac{3}{a+b+c}+\\frac{a+b+c}{2(ab+bc+ca)}\r\n\\]\r\nupon multiplying both sides by $a+b+c$ we obtain\r\n\\[\r\n3+\\frac{c}{a+b}+\\frac{a}{b+c}+\\frac{b}{c+a}\\geq 3 +\\frac{(a+b+c)^2}{2(ab+bc+ac)}\r\n\\]\r\nBoth the above inequality holds, since\r\n\\[\r\n\\sum_{cyc} \\frac{c}{a+b}=\\sum_{cyc}\\frac{c^2}{ac+bc}\\geq \\frac{(a+b+c)^2}{2(ab+bc+ac)}.\r\n\\]", + "Solution_2": "Is your solution the same?", + "Solution_3": "Unfortunately,it is. :(" +} +{ + "Tag": [ + "number theory open", + "number theory" + ], + "Problem": "This problem came up during a discussion session at a conference I attended. It looked easy and olympiad-like. But I cannot come up with a solution. I feel like I am missing something trivial.\r\n\r\nLet $ n \\geq 2$ be a positive integer and $ S$ be a subset of $ \\{ 1, 2, \\cdots, n\\minus{}1 \\}$ such that the sum of the elements of $ S$ is not divisible by $ n$. Prove that the elements of $ S$ can be ordered in a way such that no consecutive block of the ordered set $ S$ has sum divisible by $ n$.\r\n\r\nFor example, if $ n \\equal{} 8$ and $ S \\equal{} \\{ 2, 3, 5, 6, 7 \\}$, then we can order $ S$ as $ 3, 7, 2, 5, 6$, and no consecutive block of integers has sum divisible by $ 8$.", + "Solution_1": "This problem has been discussed at MO: [url]https://mathoverflow.net/q/164300[/url]" +} +{ + "Tag": [ + "geometry", + "abstract algebra", + "calculus", + "integration", + "algebra", + "function", + "domain" + ], + "Problem": "Ok, Since I'm trying to do all the commutative algebra and algebraic geometry myself as a second year undergraduate, I'll ask in this thread some small questions about theorems. Most of my previous questions which I posted in the 'problem' section were not really appropriate. So I'll ask them here\r\n\r\n\r\n[b]Question 1[/b]\r\nLet $ A$ be noetherian semilocal ring, $ I$ an ideal such that for some $ v > 0$ we have $ \\mathfrak{m}^v \\subset I \\subset \\mathfrak{m}$ where $ \\mathfrak{m}$ is the Jacobson radical. \r\n\r\nLet $ gr_I(M) \\equal{} \\bigoplus_{n \\geq 0} I^n M / I^{n \\plus{} 1} M$, then $ gr_I(M)$ is a graded module over $ gr_I(A) \\equal{} \\bigoplus_{n \\geq 0}I^n/I^{n \\plus{} 1}$. \r\nWrite $ gr_I(A) \\equal{} A'$ and $ gr_I(M) \\equal{} M'$. Then $ A_0' \\equal{} A/I$ is Artinian, and if furthermore $ I \\equal{} \\sum_{1}^{r} x_i A$, and $ \\zeta_i$ is the image of $ x_i$ in $ I/I^2$, then $ A' \\equal{} A_0'\\left[\\zeta_1,...,\\zeta_r \\right]$. \r\nIf also $ M \\equal{} \\sum_{1}^{s} A' \\bar{\\omega}_i$ where $ \\bar{\\omega}_i$ is the image of $ \\omega_i$ in $ M_0' \\equal{} M/IM$. \r\n\r\n[quote]\nThen $ l(M_n') \\equal{} l(I^nM/I^{n \\plus{} 1}M)$\n[/quote]", + "Solution_1": "[b]Question 2[/b]\r\n\r\n[quote]\nThe smallest number of elements needed to generate $ \\mathfrak{m}$ is equal to $ rank_k(\\mathfrak{m}/\\mathfrak{m}^2)$. \n[/quote]\r\nI see why this works if $ (A, \\mathfrak{m})$ is a local ring (I dropped the Noetherian condition in the theorem). Does something go wrong if I also not require $ A$ to be a local ring?", + "Solution_2": "[b]Question 3[/b]\r\n\r\nLet $ A$ be a noetherian integral domain, $ B$ a finitely generated extension ring of $ A$ which is an integral domain too. Let $ P \\in Spec \\ B$ and $ p \\equal{} P \\cap A$. Assume $ B \\equal{} A[x]$. Since we're proving a dimension inequality, we can replace (motivation follows from context) $ A$ by $ A_p$ and $ B$ by $ B_p \\equal{} A_p[x]$, so we can [u]assume[/u] $ A$ is a local ring and $ p$ its maximal ideal. Set $ I \\equal{} \\{ f(X) \\in A[X] \\mid f(x) \\equal{} 0 \\}$. If $ I\\neq 0$ then $ tr.deg_{A} B \\equal{} 0$. Because $ A$ is a subring of $ B$ we have $ I \\cap A \\equal{} 0$. Let $ K$ be the field of fraction of $ A$.\r\n\r\n[quote]\nThen $ ht \\ I \\equal{} ht \\ IK[X] \\equal{} 1$\n[/quote]\r\n\r\n:blush:", + "Solution_3": "[b]Question 4[/b]\r\n\r\nLet $ A$ be a ring and $ x \\in A$. Let $ K.(x)$ be the Koszul complex. For a complex $ C.$ of $ A$-modules we set $ C.(x)\\equal{}C. \\otimes K.(x)$. Then we can identify $ C_p(x)$ with $ C_p \\oplus C_{p\\minus{}1}$\r\n\r\nHowever the tensor product of two complexes $ K.$ and $ L_.$ is defined as $ (K \\otimes L)_n\\equal{}\\otimes_{p\\plus{}q\\equal{}n} K_p \\otimes_A L_q$.\r\n\r\nSo shouldn't we have\r\n\r\n[quote]\nThen we can identify $ C_p(x)$ with $ C_p \\otimes C_{p\\minus{}1}$\n[/quote]\r\n\r\nor where did I go wrong?", + "Solution_4": "[b]Question 5[/b]\r\n\r\nLet $ A$ be a noetherian ring, $ M$ a finite $ A$-module, $ I$ an ideal of $ A$. Suppose $ IM \\neq M$. For every $ n > 0$ we have:\r\n\r\n\"$ Ext^i_A(N,M) \\equal{} 0$ for all $ i < n$ and for some finite $ A$-module $ N$ with $ Supp(N) \\equal{} V(I)$.\"\r\n\r\nis equivalent with\r\n\r\n\"There exists an $ M$-sequence of length $ n$ contained in $ I$.\"\r\n\r\n[quote]\n[i]Proof:[/i] Assume $ I$ consists only of zero-divisors of, then there exists an associated prime $ P$ of $ M$ containing $ I$. (this is where we need the finiteness of $ M$)\n[/quote]\r\n\r\nI don't see how the finiteness condition on $ M$ is used here.", + "Solution_5": "[b]Question 6[/b]\r\n\r\nWe define the global dimension of a ring $ A$ by $ gl.dim A = \\sup \\{ proj \\ dim \\ M \\mid \\ M \\ is \\an \\ A \\ module \\}$. If $ (A, \\mathfrak{m},k)$ is a noetherian local ring \r\n[quote]\nthen $ gl.dim A = proj \\ dim_A k$\n[/quote]\r\nI'm not following here. It is known that $ proj \\ dim \\ M \\leq proj \\ dim_A \\ k$. Why is the supremum necessarily equal to $ proj \\ dim_A \\ k$?", + "Solution_6": "[b]Question 7[/b]\r\n\r\nSomething about the injective hull and completions\r\n\r\n[quote]\nLet $ (A,\\mathfrak{m})$ be a local noetherian ring. Let $ E = E_A(k)$ be the injective hull of $ k$ as $ A$-module, then $ E \\to E \\otimes_A \\hat{A}$ is surjective\n[/quote]\n\nNot following for a bit there :)\n\n[quote]\nand extends to an isomorphism\n[/quote]\r\n\r\nThat comes down to showing that $ \\hat{A}$ is faithfully flat over $ A$, I guess. So I only need a confirmation of my proof in my other thread for that." +} +{ + "Tag": [ + "algorithm", + "logarithms", + "number theory", + "Euclidean algorithm" + ], + "Problem": "Prove that the number of steps in the Euclidean algorithm is at most five times the number of digits in b, when determining $ \\gcd(a,b)$.", + "Solution_1": "[quote=\"LeFromage\"]Prove that the number of steps in the Euclidean algorithm is at most five times the number of digits in b, when determining $ \\gcd(a,b)$.[/quote]\r\n\r\nI dont have a proof but I know this result was proven by the mathematician [url=http://en.wikipedia.org/wiki/Gabriel_Lam%C3%A9]Lame[/url]. I hope that helps so you can look it up somewhere.", + "Solution_2": "How about this?\r\n\r\nLet $ b=r_{0},\\, r_{1},\\, r_{2},\\ldots$ be the successive remainder in the Euclidean algorithm applied to $ a$ and $ b$. \r\n\r\nEvery two steps reduces the remainder by at least one half: $ r_{i+2}<\\frac{1}{2}r_{i}$, for every $ i=0,\\, 1,\\, 2,\\ldots$. \r\n\r\nThus the Euclidean algorithm terminates in at most $ 2\\log_{2}(b)$ steps...and the number of steps is at most five times the number of digits in $ b$." +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "If $ a$ and $ b$ are nonnegative real numbers with $ a^2\\plus{}b^2\\equal{}4$, show that: $ \\frac{ab}{a\\plus{}b\\plus{}2} \\le \\sqrt{2}\\minus{}1$ and determine when equality occurs.", + "Solution_1": "$ \\frac {2ab}{a \\plus{} b \\plus{} 2}\\le2\\sqrt {2} \\minus{} 2$ $ \\implies$ $ 2ab \\equal{} (a \\plus{} b)^2 \\minus{} 4$ $ \\implies$ \r\n$ a \\plus{} b \\minus{} 2\\le2\\sqrt {2} \\minus{} 2$ This is cauchy schwarz inequality :wink:", + "Solution_2": "$\\color{blue} \\boxed{\\textbf{SOLUTION}}$\n\nWe have, \n$$\\frac{ab}{a+b+2}=\\frac{(a+b)^2 -4}{2} \\times \\frac{1}{a+b+2}=\\frac{(a+b+2)(a+b-2)}{2} \\times \\frac{1}{a+b+2}=\\frac{a+b-2}{2}=\\frac{a+b}{2} -1$$\n\nSo, We need to show,\n$$\\frac{a+b}{2} -1 \\le \\sqrt2 -1 \\implies (a+b)^2 \\le 8 \\implies ab \\le 2$$\n\nNow, by $\\textbf{AM-GM Inequality}$ \n$$ab \\le \\frac{(a+b)^2}{4} = 1+\\frac{ab}{2} \\implies \\frac{ab}{2} \\le 1 \\implies ab \\le 2$$\n\nEquality hold for $ab=2 \\blacksquare$", + "Solution_3": "[quote=moldovan]If $ a$ and $ b$ are nonnegative real numbers with $ a^2\\plus{}b^2\\equal{}4$, show that: $ \\frac{ab}{a\\plus{}b\\plus{}2} \\le \\sqrt{2}\\minus{}1$ and determine when equality occurs.[/quote]\n\nBro why did I suffer on this for 10 minutes? \n\nNote that by AM GM $ab\\leq 2$. So $(a+b)^2=4+2ab\\leq 8$ so $a+b\\leq 2\\sqrt 2$. Now $$\\frac{ab}{a+b+2}=\\frac{(a+b)^2-4}{2(a+b+2)}=\\frac 12(a+b-2)\\leq \\frac 12(2\\sqrt 2-2)=\\sqrt 2-1$$", + "Solution_4": "Let $ a , b$ are real numbers with $ a^2+b^2=4 $ and $ ab\\neq 0.$ Show that$$ \\frac{ab}{a+b+2} \\le \\sqrt{2}-1$$" +} +{ + "Tag": [], + "Problem": "I'm looking for PDF whose topic is Functional Equations.\r\n\r\nDoes anyone knows?", + "Solution_1": "Go to my page and see Download section //http://trungtuan.wordpress.com/", + "Solution_2": "[quote=\"N.T.TUAN\"]Go to my page and see Download section //http://trungtuan.wordpress.com/[/quote]\r\n\r\nYour link doesnt work :wink:", + "Solution_3": "No, it is good, remove ''//'' . :wink:", + "Solution_4": "To\u3000N.T.TUAN\r\n\r\nDownload section is disapperaed.", + "Solution_5": "the link TUAN was refering to is:\r\n[url]http://trungtuan.wordpress.com/[/url]", + "Solution_6": "another link pease maybe rapidshare", + "Solution_7": "what's wrong with that link? :huh:" +} +{ + "Tag": [ + "Support" + ], + "Problem": "I know that badges are only awarded for passes or mastering a section. However, when i lost my 2 badges, i couldn't help thinking: Is it possible to get my badges back? Anybody who is willing to answer my question will be deeply appreciated. By the way, the two badges [b]were \"Counting Numbers in Lists\" and \"Venn Diagrams\"...\r\n\r\nThanks in advance!! :o :(", + "Solution_1": "Yah, u lose badges by \"un-mastering\" them, aka, getting them wrong, so they take away ur badges. If u do the problems good, and get alot correct of venn diagram, and the other topic, u get the badges back.\r\n\r\nlol, i lost venn diagram 3 times, but thankfully, i think i have taht badge right now...... :)", + "Solution_2": "From the [url=http://www.artofproblemsolving.com/Alcumus/FAQ.php]FAQ[/url]:\r\n\r\n[b]How come I can no longer see badges I use to be able to see?[/b]\r\nYou will only be able to see badges from subjects in which you have passed or mastered. If you begin to answer problems incorrectly in a subject or related subject from which you have previously passed or mastered, or others start answering problems correctly that you previously answered incorrectly, it may be determined that you have no longer passed or mastered a subject. At that time you will no longer be able to see the badges. Once you have sufficiently answered problems in the subject and related subjects, you will again be flagged as \"passed\" or \"mastered\" in that subject and the badge will appear.", + "Solution_3": "ok, thanks everyone... i just got my badges back through a couple more correct answers. Now i understand how it works... Thanks everyone!!" +} +{ + "Tag": [ + "geometry", + "inequalities", + "geometry unsolved" + ], + "Problem": "Prove that for any point $P$ in the interior of a triangle $ABC$, the sum $ PA\\cdot a + PB\\cdot b +PC\\cdot c$ is bigger or equal than $4(ABC)$.\n\n[i]Remark.[/i] Here, $(ABC)$ denotes the area of triangle $ABC$.", + "Solution_1": "you have posted the same problem here:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=198639[/url]\r\n\r\nposting the same problem again doesn't help you get the solution,so stop it. :|", + "Solution_2": "I know i posted the problem mate, but because it turned out to be in spanish i think people stop givin the solution, so please give the solution anyone that can.\r\n\r\n Thank you", + "Solution_3": "posting spanish in the other topic was a wrong thing to do,anyway please don't double post again it's spamming.", + "Solution_4": "i wont do it again, sorry, \r\n\r\n\r\n\r\n Does anyone have the solution?", + "Solution_5": "let $ PA \\equal{} x,PB \\equal{} y,PC \\equal{} z$ now let $ m,n,p$ denote the distances from point $ P$ to the sides $ BC,CA,AB$ respectively,now we have to prove that:\r\n\r\n$ ax \\plus{} by \\plus{} cz\\geq 4S$\r\n\r\nwhere $ S$ denotes the area of $ \\triangle ABC$.\r\n\r\nnow let $ H$ be a point on side $ BC$ such that $ AH\\perp BC$ and let $ h_a \\equal{} AH$,now we obviously have:\r\n\r\n$ x \\plus{} m\\geq h_a$\r\n\r\n$ \\Rightarrow a(x \\plus{} m)\\geq ah_a \\equal{} 2S$\r\n\r\n$ \\Rightarrow ax\\geq 2S \\minus{} am$\r\n\r\nin the same way we conclude that:\r\n\r\n$ by\\geq 2S \\minus{} bn$\r\n\r\nand\r\n\r\n$ cz\\geq 2S \\minus{} cp$\r\n\r\nthus:\r\n\r\n$ ax \\plus{} by \\plus{} cz\\geq 6S \\minus{} (am \\plus{} bn \\plus{} cp)$ (*)\r\n\r\nnow note that in triangle $ \\triangle PBC$ we have:\r\n\r\n$ \\frac 12.ma \\equal{} S_{\\triangle PBC}\\Rightarrow am \\equal{} 2S_{\\triangle PBC}$\r\n\r\nsimilary:\r\n\r\n$ bn \\equal{} 2S_{\\triangle PCA}$\r\n\r\nand\r\n\r\n$ cp \\equal{} 2S_{\\triangle PAB}$\r\n\r\nso according to (*) we get that:\r\n\r\n$ ax \\plus{} by \\plus{} cz\\geq 6S \\minus{} 2\\left(S_{\\triangle PBC} \\plus{} S_{\\triangle PCA} \\plus{} S_{\\triangle PAB}\\right)$\r\n\r\n$ \\Rightarrow ax \\plus{} by \\plus{} ca\\geq 6S \\minus{} 2S \\equal{} 4S$\r\n\r\nQED", + "Solution_6": "Please do not answer posts like this. This problem is part of a homework that Chingonkan999 must do alone. The homework is part of Mexico's training program.\r\n\r\nMaybe mexican trainers haven't noticed, but is a little illegal to do what chingonkan999 did.", + "Solution_7": "I am sorry paqui but this is not illegal at all,\r\n\r\n I already told the main mexican trainer and he even said i could ask for help from the trainer that gave us the homework.\r\n \r\n It is not the first time i upload a problem, it is not even the first time someone uploads a problem of the homework. Even you uploaded a problem of the homework, the one of the 21agon so please do not say things like that.\r\n\r\n Thankyou", + "Solution_8": "It's not like this is a new problem or anything...I'm sure it could be found a book or something.", + "Solution_9": "Let $ M,N,L$ be the midpoints of $ BC,CA,AB.$ In the quadrangles $ PLAN,$ $PMBL,$ $PNCM,$ we have the inequalities:\n\n$ [PLAN] \\le \\frac {PA \\cdot NL}{2} \\equal{} \\frac {PA \\cdot BC}{4}$ \n\n$[PMBL] \\le \\frac {PB \\cdot LM}{2} \\equal{} \\frac {PB \\cdot CA}{4}$ \n\n$[PNCM] \\le \\frac {PC \\cdot MN}{2} \\equal{} \\frac {PC \\cdot AB}{4}$\n\nAdding the 3 previous inequalities gives $ PA \\cdot BC \\plus{} PB \\cdot CA \\plus{} PC \\cdot AB \\ge 4[ABC]$" +} +{ + "Tag": [ + "inequalities", + "quadratics" + ], + "Problem": "This was just a problem which I thought was the hardest question in our half semester tests over the last 10 years. A bit easy for year 10 though but here it is: :D \r\n\r\n$2x^2 + 6xy +10y^2 -2x+6y+10=0$\r\n\r\nProve that there are no real solutions for $(x, y)$", + "Solution_1": "[hide]\nrewrite as $(x+3y)^2+(x-1)^2+(y+3)^2=0$ that lead us to the system\n\n$x+3y=0$\n$x-1=0$\n$y+3=0$\n\nthat has not real solutions[/hide]", + "Solution_2": "[quote=\"hydro\"][hide]\nrewrite as $(x+3y)^2+(x-1)^2+(y+3)^2=0$ that lead us to the system\n\n$x+3y=0$\n$x-1=0$\n$y+3=0$\n\nthat has not real solutions[/hide][/quote]\r\n\r\nWhy does each unit have to equal zero for the equation to equal zero? Couldn't one unit equal -4, another -4, and the last 8 (as an example)?", + "Solution_3": "[quote=\"surge\"][quote=\"hydro\"][hide]\nrewrite as $(x+3y)^2+(x-1)^2+(y+3)^2=0$ that lead us to the system\n\n$x+3y=0$\n$x-1=0$\n$y+3=0$\n\nthat has not real solutions[/hide][/quote]\n\nWhy does each unit have to equal zero for the equation to equal zero? Couldn't one unit equal -4, another -4, and the last 8 (as an example)?[/quote]\r\n\r\nmmmh... i think that a square has to be positive, and also the sum of three squares must be $\\geq 0$ in $\\mathbb {R}$", + "Solution_4": "[quote=\"sen\"]This was just a problem which I thought was the hardest question in our half semester tests over the last 10 years. A bit easy for year 10 though but here it is: :D \n\n$2x^2 + 6xy +10y^2 -2x+6y+10=0$\n\nProve that there are no real solutions for $(x, y)$[/quote]\r\n\r\n$(3y+x)^2+(x-1)^2+(y+3)^2=0$\r\n\r\nnow by the trivial inequality, each square has to be greater than or equal to 0, but since there is equality, each square is equal to 0, and there is no solution to that system that results from the squares being 0", + "Solution_5": "or you can just be all like hmmmmm quadratic equation!" +} +{ + "Tag": [ + "limit", + "algebra proposed", + "algebra" + ], + "Problem": "Let two sequence (x_n),(y_n) are defined by\r\n$ x_0\\equal{}\\frac{2007}{2008}, y_0\\equal{}\\frac{2008}{2009},$ \r\n$ x_{n\\plus{}1}\\equal{}x_n\\plus{}\\frac{1}{y_n}, y_{n\\plus{}1}\\equal{}y_n\\plus{}\\frac{1}{x_n}.$\r\nFind the limit of $ \\frac{1}{x_n\\plus{}y_n}$", + "Solution_1": "Denote by $ A_n\\equal{}x_ny_n \\Rightarrow A_{n\\plus{}1}\\equal{}A_n\\plus{}2\\plus{}\\frac {1} {A_n} \\Rightarrow \\displaystyle\\lim_{x\\to\\infty} A_n\\equal{}\\infty$\r\n\r\nSince $ 0<\\frac {1} {x_n\\plus{}y_n}$ and $ \\frac {1} {x_n\\plus{}y_n} \\le \\frac {1} {2\\sqrt{x_ny_n}} \\rightarrow 0 \\Rightarrow \\displaystyle\\lim_{x\\to\\infty} \\frac {1} {x_n\\plus{}y_n}\\equal{}0$" +} +{ + "Tag": [ + "AMC", + "AIME", + "MATHCOUNTS" + ], + "Problem": "Check this out...I'm still working on it (obviously) but I was looking for old stuff, and I couldn't find it, so I thought it would be useful to compile it...\r\n\r\n\r\nhttp://mathjunk.googlepages.com\r\n\r\n\r\nIt was designed for private use at first, but I decided to share it with AoPSers as well.", + "Solution_1": "I also started one\r\n\r\nhttp://theani.isgreat.org/tmp/contests.php\r\n\r\nbut its going to change", + "Solution_2": "i found the first site very helpful", + "Solution_3": "[quote=\"star99\"]i found the first site very helpful[/quote]\r\n\r\n\r\nI second it!\r\n\r\n\r\nGO MYSMARTMOUTH!!!! WOOO!!!! :coolspeak:", + "Solution_4": "wow nice! good compilation of AMC/AIME/Mathcounts problmes", + "Solution_5": "I like the AMC 10 ones. Not enough online on AoPS and other websites. :lol:\r\n\r\n\r\nthanks", + "Solution_6": "The layout sucked, so I changed it, and this way I will have much more storage space too:\r\n\r\nhttp://mathjunk.googlepages.com\r\n\r\nI added a bunch of stuff, and the only thing missing is the 2000 Mandelbrot, which is lost somewhere on my hard drive.", + "Solution_7": "good work, sean :)", + "Solution_8": "I want some amc 12s please. :lol:", + "Solution_9": "Sorry this is kind of late, but could you maybe PM me the password for the AIME and Mathcounts pages? \r\nThanks." +} +{ + "Tag": [ + "modular arithmetic", + "algebra proposed", + "algebra" + ], + "Problem": "Does there exist irrational $x,y>1$ that for each $m,n\\in\\mathbb Z$ \\[[x^{m}]\\neq[y^{n}]\\]", + "Solution_1": "I have already posted a solution for this one: take numbers of the form $a+\\sqrt{a^{2}-1}$ for both numbers $x,y$. Then it's easy.", + "Solution_2": "[i]My Solution :[/i]\r\nPut $x=2+\\sqrt 2$ and $y=\\frac{3+\\sqrt 5}2$. Then for each $m,n\\in\\mathbb N$ we can prove easily that $[x^{m}]\\equiv-1\\pmod4$ and $[y^{n}]\\equiv1,2\\pmod4$", + "Solution_3": "harazi,\r\n\r\nState what type of number \"a\" is first, before typing beyond about the form it is inside of.\r\n\r\nThen it's \"easy?\" No, show the main work for that so-called\r\n\"easy\" step, because you're incomplete in your solution, regardless of your claiming to have posted a solution before.", + "Solution_4": "[quote=\"Omid Hatami\"][i]My Solution :[/i]\nPut $x=2+\\sqrt 2$ and $y=\\frac{3+\\sqrt 5}2$. Then for each $m,n\\in\\mathbb N$ we can prove easily that $[x^{m}]\\equiv-1\\pmod4$ and $[y^{n}]\\equiv1,2\\pmod4$[/quote]\r\n\r\nThe question is: how did you manage to find those $x;y$?", + "Solution_5": "I knew that $(\\frac{3+\\sqrt5}2)^{n}+(\\frac{3-\\sqrt5}2)^{n}-2$ is always perfect square. The second one is obvious.", + "Solution_6": "To a cute angle: don't be frustrated! I said I already posted a complete solution, I'm not going to write it again." +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "Let $a,b,c$ are sides of triangle such that $a+b+c=1$. Prove that:\r\n\\[\\frac{1}{1-2(a^{2}+b^{2}+c^{2})}\\ge \\frac{a^{2}+b^{2}+c^{2}}{a^{2}b+b^{2}c+c^{2}a}\\]", + "Solution_1": "Let a=y+z,b=y+x,c=x+z,then\r\n\r\n(a+b+c)/((a+b+c)^2-2*a^2-2*b^2-2*c^2)-(a^2+b^2+c^2)/(a^2*b+b^2*c+c^2*a)>=0\r\n\r\n<==>\r\n\r\na^3*b+a*b^2*c-3*c^2*a^2-3*a^2*b^2+b^3*c+b*c^2*a+c*a^2*b\r\n\r\n-3*b^2*c^2+c^3*a-a^4+2*a^3*c+2*b^3*a-b^4+2*c^3*b-c^4>=0\r\n\r\n<==>\r\n\r\n[y^2*(y-x)*(y-z)+x^2*(x-y)*(x-z)+z^2*(z-y)*(z-x)]\r\n\r\n+[x*z*(y-x)*(y-z)+y*z*(x-y)*(x-z)+y*x*(z-y)*(z-x)]\r\n\r\n+[y*z*(y-x)^2+x*y*(x-z)^2+z*x*(z-y)^2]>=0.\r\n\r\nThe inequality holds." +} +{ + "Tag": [ + "function", + "algebra", + "domain", + "parameterization", + "algebra unsolved" + ], + "Problem": "find all function such that \r\n$ (\\forall( x,y,n) \\in \\mathbb{R}^2 \\mathbb{N}^*)f(x)\\plus{}f(y^n)\\equal{}f(x)f(y)^n$", + "Solution_1": "[quote=\"mathmen\"]find all function such that \n$ (\\forall( x,y,n) \\in \\mathbb{R}^2 \\mathbb{N}^*)f(x) \\plus{} f(y^n) \\equal{} f(x)f(y)^n$[/quote]\r\nHi, I will not solve this problem. I just want to pass to tell you that, please post the problem at the correct forum.\r\n\r\nThis one should be posted at Algebra Section.", + "Solution_2": "think you :arrow: :arrow:", + "Solution_3": "$ f(x) \\plus{} f(y^n) \\equal{} f(x)f(y)^n \\;\\;\\;\\forall x,y \\in \\mathbb{R}, \\forall n \\in \\mathbb{N}^*\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}(1)$\r\n$ x \\equal{} y \\equal{} 1, n \\equal{} 1$ in $ (1) \\Rightarrow 2f(1) \\equal{} f(1)^2 \\Rightarrow f(1) \\equal{} 0$ or $ 2$.\r\n$ x \\equal{} y \\equal{} 1, n \\equal{} 2$ in $ (1) \\Rightarrow 2f(1) \\equal{} f(1)^3 \\Rightarrow f(1) \\equal{} 0$ or $ \\sqrt{2}$ or $ \\minus{}\\sqrt{2}$.\r\nHence, $ f(1) \\equal{} 0$. \r\nSetting $ x \\equal{} 1$ in $ (1)$, we get\r\n$ f(x) \\plus{} 0 \\equal{} 0 \\Rightarrow f(x) \\equal{} 0\\;\\;\\; \\forall x \\in \\mathbb{R}$.\r\n\r\nI would like to point out that we only needed to suppose that $ n$ assumed values $ 1$ and $ 2$.", + "Solution_4": "Take x=y=a and n=1 to get $ 2f(a)\\equal{}f(a)^2$ so $ f(a)(f(a)\\minus{}2)\\equal{}0$ so $ f(a)\\equal{}0 or 2\\forall a\\in\\mathbb{R}$. If there exists a b such that $ f(b)\\equal{}0$ than take n=1 so $ \\forall x\\in\\mathbb{R} f(x)\\plus{}f(b)\\equal{}f(x)f(b)$, so $ f(x)\\equal{}0\\forall x\\in\\mathbb{R}$. If no such b exists than $ f(x)\\equal{}2\\forall x\\in\\mathbb{R}$. Using n=2 easily disproves this so the only solution is $ f(x)\\equal{}0\\forall x\\in\\mathbb{R}$.", + "Solution_5": "[quote=\"mathmen\"]find all function such that \n$ (\\forall( x,y,n) \\in \\mathbb{R}^2 \\mathbb{N}^*)f(x) \\plus{} f(y^n) \\equal{} f(x)f(y)^n$[/quote]\r\nIt is not clear if $ (0,0,n)$ are included or not...I suppose they are included\r\n\r\nIf $ y \\equal{} 0$ $ \\implies f(x) \\plus{} f(0) \\equal{} f(x)f(0)^n \\implies f(x) \\equal{} \\frac {f(0)}{f(0)^n \\minus{} 1}$ with $ f(0)\\ne1$\r\n\r\nIf $ f(0) \\equal{} 1$ $ \\implies f(x) \\plus{} 1 \\equal{} f(x)$ that is impossible\r\n\r\nIf $ x \\equal{} 0,y \\equal{} 0$ $ \\implies 2f(0) \\equal{} f(0)^{n \\plus{} 1}$ so either $ 1) f(0) \\equal{} 0$ or $ 2) f(0) \\equal{} \\sqrt [n]{2}$ $ n\\in\\mathbb{N}^*$\r\n\r\ncase $ 1) \\implies f(x) \\equal{} 0, \\forall x \\in \\mathbb{R}$ and this is already found solution\r\n\r\ncase $ 2) \\implies f(x) \\equal{} \\sqrt [n]{2}, \\forall x \\in \\mathbb{R}$ and $ n\\in\\mathbb{N}^*$\r\n\r\nIt is simple check that last is indeed a solution of original equation. The problem is: what is the domain of this function? We may intend $ n$ as parameter and so we have infinite many constant solutions. Otherwise, we have $ f: \\mathbb{N}^* \\rightarrow \\mathbb{R}$ that is NOT (post-edited) a solution (in the statement is not clear what function's domains are searched or not)." +} +{ + "Tag": [ + "geometry", + "symmetry" + ], + "Problem": "1. In each of the following pairs, select the compound with the highest boiling point, always justifying your answer:\r\n\r\n(a) $ \\ce{SiH4}$ or $ \\ce{GeH4}$\r\n\r\n(b) Ethyl fluoride or ethyl bromide\r\n\r\n(c) $ \\ce{Br2}$ or $ \\ce{I2}$\r\n\r\n(d) $ \\ce{BF3}$ or $ \\ce{ClF3}$\r\n\r\n(e) $ \\ce{SF4}$ or $ \\ce{CF4}$\r\n\r\n(f) $ \\ce{CHF3}$ or $ \\ce{CF4}$\r\n\r\n(g) Ne or Xe\r\n\r\n(h) $ \\ce{SF4}$ or $ \\ce{SF6}$\r\n\r\n(i) $ \\ce{PF3}$ or $ \\ce{PCl3}$\r\n\r\n(j) $ \\ce{SO2}$ or $ \\ce{CO2}$\r\n\r\n(k) 1,1-Dichloroethene or trans-1,2-dichloroethene\r\n\r\n(l) Ethyl ether or ethyl sulphide\r\n\r\n(m) o-Dichlorobenzene or p-dichlorobenzene\r\n\r\n(n) $ \\ce{BF3}$ or $ \\ce{BCl3}$\r\n\r\n(o) Phenol or thiophenol\r\n\r\n(p) cis-1,2-Dichloroethene or trans-1,2-dichloroethene\r\n\r\n(q) Trimethylamine or ethyl methyl amine\r\n\r\n(r) trans-1,2-Dichloroethene or trans-1,2-dibromoethene", + "Solution_1": "2. In which of the following substances exist hydrogen bonds between their molecules?\r\n\r\n(a) Methylamine\r\n\r\n(b) HF\r\n\r\n(c) Methyl ether\r\n\r\n(d) Acetic acid\r\n\r\n(e) Ammonia\r\n\r\n(f) HBr\r\n\r\n(g) Methanol\r\n\r\n(h) $ \\ce{D2O}$\r\n\r\n(i) Ethanol\r\n\r\n(j) Acetaldehyde\r\n\r\n(k) Phosphane\r\n\r\n(l) $ \\ce{HClO}$\r\n\r\n(m) Methane\r\n\r\n(n) Phosphoric acid", + "Solution_2": "3. Explain each of the following experimental facts:\r\n\r\n(a) The boiling point of n-pentane is 36\u00baC while the boiling point of its isomer neopentane (tetramethylmethane) is 10\u00baC.\r\n\r\n(b) Methanol has a higher vapour pressure than $ \\ce{CH3SH}$.\r\n\r\n(c) At ambient T and P, n-pentane is liquid while propane is a gas.\r\n\r\n(d) In general, aldehydes and ketones have lower the boiling points than alcohols with similar molecular weights.\r\n\r\n(e) At room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid.\r\n\r\n(f) Ethyl ether has a higher vapour pressure than water.\r\n\r\n(g) Ammonia has a higher boiling point (-33\u00baC) than phosphane (-87\u00baC).\r\n\r\n(h) o-Hydroxybenzaldehyde has a lower boiling point than p-hydroxybenzaldehyde.\r\n\r\n\r\n4. Without consulting tables, rank the following alkanes by increasing boiling point:\r\n\r\n(a) 3,3-Dimethylpentane\r\n(b) n-Heptane\r\n(c) 2-Methylheptane\r\n(d) n-Pentane\r\n(e) 2-Methylhexane\r\n\r\n\r\n5. Why is pure hydrogen peroxide a substance more viscous, with a lower vapour pressure, and with a higher boiling point (152\u00baC) compared to water?\r\n\r\n\r\n6. Without consulting tables, rank the following compounds by increasing boiling point:\r\n\r\n(a) Butan-2-one\r\n(b) Acetone\r\n(c) Hexan-2-one\r\n(d) 4-Methylpentan-2-one\r\n(e) n-Butyl phenyl ketone\r\n(f) 3,3-Dimethylbutan-2-one\r\n\r\n\r\n7. Without consulting tables, rank the following compounds by increasing boiling point:\r\n\r\n(a) Ethylamine\r\n(b) Methane\r\n(c) Chloroethane\r\n(d) Sodium acetate\r\n(e) Ethanol\r\n(f) Acetic acid\r\n(g) Ethane\r\n\r\n\r\n8. In each of the following pairs, select the compound with the highest melting point:\r\n\r\n(a) tert-Butanol or n-butanol\r\n\r\n(b) Methyl propanoate or sodium propanoate\r\n\r\n(c) Benzene or toluene\r\n\r\n(d) o-Dichlorobenzene or p-dichlorobenzene\r\n\r\n(e) Propane or cyclopropane", + "Solution_3": "Im just going to answer number 2 because it looks easiest:\r\nMy guesses are:\r\n\r\na) Methylamine \r\n(b) HF \r\n(d) Acetic acid \r\n(e) Ammonia\r\n(g) Methanol \r\n(i) Ethanol \r\n(n) Phosphoric acid", + "Solution_4": "There is one compound missing there.", + "Solution_5": "Is it HOCl?", + "Solution_6": "Is it (h) $ \\ce{D_{2}O}$ ?", + "Solution_7": "HOCl, most likely.\r\n\r\nthe molecule looks something like Cl--O--H\r\nHOCl is a weak acid.\r\nHydrogen bondage! (because of the hydroxyl group)", + "Solution_8": "Yes, hypochlorous acid was the missing compound. What about the other questions?", + "Solution_9": "Let me attempt #1...\r\n1.\r\na. Germane (?) $ \\mathrm{GeH_4}$\r\nBoth molecules are nonpolar, so the molecule with more surface area and more electrons available for dispersion will have the higher boiling point. Germanium is larger, heavier, and has more electrons than silicon, being lower down in the periodic table. So, germane has the higher boiling point.\r\n\r\nb. Ethyl bromide.\r\nI would say that the halogens are an anomaly when it comes to organic compounds featuring halogens. Rather than the polarity due to the electronegativity difference, I would think that the ethyl fluoride would be more polar and thus have a higher boiling point, but that is not the case. Size (dispersion) matters more.\r\n\r\nc. $ \\mathrm{I_2}$\r\nIodine is much more heavy than bromine. Also, one should recall that bromine is a liquid and iodine is a solid at room temperature.\r\n\r\nd. $ \\mathrm{ClF_3}$\r\nThe chlorine molecule has a dipole moment arising from its 3D structure. The difference in electronegativity between Cl and F, in addition to the lone pair on the chlorine atom, results in a dipol,e. BF3 is a planar molecule.\r\n\r\n...To Be Continued", + "Solution_10": "Your answers are correct, although the explanations could be better.", + "Solution_11": "Answers to number 3:\r\n\r\na) The neopentane isomer does not have as much contact between molecules as the n-pentane. Hence neopentane experiences smaller london dispersion forces, thereby causing a smaller boiling point.\r\n\r\nb) This one I would think that methanol would have a lower Pvap than methyl mercaptan as the methanol H-bonds, hence the greater IMF's cause more methanol in to remain in solution..... not sure though, Methyl mercaptan could be experiencing a stronger dipole dipole...\r\n\r\nc)n-pentane is a larger by a whole ethyl group, hence it has greater london dispersion forces.\r\n\r\nd) alcohols have H-bonds thus they take more energy to break, than aldehydes and ketones, which do not have the H-bonding ability\r\n\r\ne) F and Cl as smaller than Br and I, hence they experience lesser vanderwaals forces. Similarly, Br is smaller than I, thus the same applies in this case.\r\n\r\nf) water hydrogen bonds, ethers do not... hence water will have a greater tendancy to remain as a liquid.\r\n\r\ng)Ammonia hydrogen bonds.... phosphane does not (kind of repetitive...)", + "Solution_12": "How about part (h)?", + "Solution_13": "is part h) due to the symmetry of the para compound over the ortho one? :maybe:", + "Solution_14": "No, that's not the reason.", + "Solution_15": "My guesses:\r\n\r\n[quote]1. In each of the following pairs, select the compound with the highest boiling point, always justifying your answer:\n\n(a) $ \\ce{SiH4}$ or $ \\ce{GeH4}$[/quote]\n\nthe germanium compound is heavier with more loosely bound electrons that can wander around and induce temporary dipoles and so forth\n\n[quote](b) Ethyl fluoride or ethyl bromide[/quote]\n\nagain, the bromide is much bigger\n\n[quote](c) $ \\ce{Br2}$ or $ \\ce{I2}$[/quote]\n\nnow the iodine is bigger\n\n[quote](d) $ \\ce{BF3}$ or $ \\ce{ClF3}$[/quote]\nthe boron compound is planar and symmetric with no overall dipole moment, so the second, with a trigonal bipyramidal (?) geometry and thus a permanent dipole, should have stronger intermolecular forces\n\n[quote](e) $ \\ce{SF4}$ or $ \\ce{CF4}$[/quote]\ncarbon tetrafluoride is tetrahedral with obvious symmetry, so now the first, with a see-saw shape, wins\n\n[quote](f) $ \\ce{CHF3}$ or $ \\ce{CF4}$[/quote]\nit looks like now the second has no dipole moment and the first does, so the first wins\n\n[quote](g) Ne or Xe[/quote]\nxenon is bigger\n\n[quote](h) $ \\ce{SF4}$ or $ \\ce{SF6}$[/quote]\nI'm not sure, but I think the latter will now have a symmetrical structure and no dipole moment\n\n[quote](i) $ \\ce{PF3}$ or $ \\ce{PCl3}$[/quote]\nthe latter should be bigger\n\n[quote](j) $ \\ce{SO2}$ or $ \\ce{CO2}$[/quote]\nthe first has a dipole moment\n\n[quote](k) 1,1-Dichloroethene or trans-1,2-dichloroethene[/quote]\nthe first has a dipole moment\n\n[quote](l) Ethyl ether or ethyl sulphide[/quote]\n\nthe sulfur compound is bigger\n\n[quote](m) o-Dichlorobenzene or p-dichlorobenzene[/quote]\nthe first has a dipole\n\n[quote](n) $ \\ce{BF3}$ or $ \\ce{BCl3}$[/quote]\nthe latter is larger\n\n[quote](o) Phenol or thiophenol[/quote]\nthe first has hydrogen bonding\n\n[quote](p) cis-1,2-Dichloroethene or trans-1,2-dichloroethene[/quote]\nthe first has a dipole moment\n\n[quote](q) Trimethylamine or ethyl methyl amine[/quote]\nthe latter has hydrogen bonding\n\n[quote](r) trans-1,2-Dichloroethene or trans-1,2-dibromoethene[/quote]\r\nthe latter is larger", + "Solution_16": "more guesses:\r\n\r\n\r\n[quote](h) o-Hydroxybenzaldehyde has a lower boiling point than p-hydroxybenzaldehyde.[/quote]\n\nso a lot of the intermolecular forces will be arising from the strong dipole associated with the O-H bond in the hydroxyl. my guess is that with the para molecule, it's easier for the positively charged H to attract with the negatively charged O of the aldehyde group of another molecule (and then the other molecule's hydroxyl group in turn stabilizes someone else, etc), whereas all of this would be much more scrunched and sterically hindered with the ortho molecule\n\n\n[quote]4. Rank the following alkanes by increasing boiling point:\n\n(a) 3,3-Dimethylpentane\n(b) n-Heptane\n(c) 2-Methylheptane\n(d) n-Pentane\n(e) 2-Methylhexane[/quote]\n\nI would guess the pentane, then 3,3-dimethylpentane, then 2-methylhexane, then heptane, and finally 2-methylheptane.\n\n\n[quote]5. Why is pure hydrogen peroxide a substance more viscous, with a lower vapour pressure, and with a higher boiling point (152\u00baC) compared to water?[/quote]\n\nI'm not sure I get this. If we really did just have hydrogen peroxide, then you could say it has twice as much hydrogen bonding as water or something like that. But the hydrogen peroxide would just decompose into water and oxygen gas? \n\n[quote]Rank the following compounds by increasing boiling point:\n\n(a) Ethylamine\n(b) Methane\n(c) Chloroethane\n(d) Sodium acetate\n(e) Ethanol\n(f) Acetic acid\n(g) Ethane[/quote]\r\n\r\nso methane has the lowest boiling point, then ethane, then chloroethane. Then ethylamine starts to have hydrogen bonding, then ethanol, then acetic acid. Sodium acetate is ionic and should have huge mp/bp.", + "Solution_17": "Your explanation for (h) is still not correct. The rest looks ok.", + "Solution_18": "in ortho hydroxy benzaldehyde intramolecular hydrogen bonding takes place(chelation). that effectively reduces the surface area and also the ability to hydrogen bond intermoleuclarly. in case of para, there is no suchprib to hydrogen bond intermolecularly and since boiling point depepnds on intermolecular forces, and NOT on intramolecualr forces, para has greater boiling point.", + "Solution_19": "Yes, that's correct, but chelation does not have nothing to do with this." +} +{ + "Tag": [ + "abstract algebra" + ], + "Problem": "A jar contains 9000 kernels of corn, and 3 of these kernels placed side by side would measure one inch. How many yards would they reach if all 9000 were placed side by side? Express your answer as a mixed number.", + "Solution_1": "9000/3=3000\r\n(3000/12)/3=3000/36\r\n\r\n3000/36= [b]83 1/3[/b]", + "Solution_2": "First off, There are 36 inches in a yard.\r\n\r\n3 kernels = 1 inch\r\n3 kernels = 1/36 of a yard\r\n1 kernel = 1/108 of a yard\r\n\r\nYour jar has 9000 kernel, and you now know that each kernel is 1/108 of a yard.\r\n\r\n9000 x (1/108) = 83 1/3\r\n\r\nYou get 83 yards and 1/3 of a yard.", + "Solution_3": "Check your calculations. The answer is $ \\boxed{83\\frac13}$, as covertcognomen stated." +} +{ + "Tag": [], + "Problem": "How many positive integers x there are such that x and x+99 are both perfects squares?", + "Solution_1": "[hide]$x + 99 = k^2$ and $x = h^2$ $\\Longrightarrow$ $(k - h)(k + h) = 99 = 3^2 \\cdot 11$. \nNow it's just counting cases.[/hide]", + "Solution_2": "I got 1: x = 1;\r\nedit:\r\ngot that wrong...", + "Solution_3": "[hide]Using $(k-h)(k+h)=99$, we know $k-h$ and $k+h$ must be factors of 99. The only pairs of factors of 99 are 1 and 99, 3 and 33, and 9 and 11. So there are 3 values of x: 2401, 225, and 1.[/hide]" +} +{ + "Tag": [ + "function", + "number theory proposed", + "number theory" + ], + "Problem": "I think I have proved that:\r\n\r\n$ \\zeta(3)=\\frac{\\pi^{6}}{\\pi^{6}-945}\\sum_{f: \\mu(f)=0}^{\\infty}\\frac{1}{f^{3}}$ \r\n\r\n$ \\mu(n)$ is the M\u00f6bius function", + "Solution_1": "As every integer $ n$ can be written as $ n= f \\cdot s^{2}$ with $ f$ squarefree in a unique way, we get that $ \\zeta(k) = \\sum_{n=1}^\\infty \\frac{1}{n^{k}}= \\sum_{f \\in F}\\frac{1}{f^{k}}\\cdot \\sum_{s=1}^\\infty \\frac{1}{s^{2k}}= F(k) \\cdot \\zeta(2k)$.\r\nHere $ F$ is the set of squarefree numbers and $ F(k)=\\sum_{f \\in F}\\frac{1}{f^{k}}$. And if we define $ G(k) = \\zeta(k)-F(k) = \\sum_{g \\not\\in F}\\frac{1}{g^{k}}$, we get that $ \\zeta(k) = (\\zeta(k)-G(k))\\zeta(2k)$, thus $ \\zeta(k) = \\frac{G(k)}{\\zeta(2k)-1}$.\r\n\r\nUsing that we know $ \\zeta(2k)=-\\frac{(2 \\pi i)^{2k}B_{2k}}{2 \\cdot (2k)!}$, we can thus describe $ \\zeta(k)$ the way you said (especially for $ k=3$).", + "Solution_2": "That's a much more elegant way to show it than I did and you even put it in terms of any s. Thanks!" +} +{ + "Tag": [ + "MATHCOUNTS" + ], + "Problem": "Why does it say that the All Time Greatest Mathcounts Problems is out of stock.......permanently??? \r\nhttp://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=6 :wallbash:", + "Solution_1": "Maybe the MATHCOUNTS people aren't selling them. Or the AoPS staff don't get a profit from those books. There are several reasons.\r\n\r\nWhy don't you PM an admin?", + "Solution_2": "http://secure.sportsawardsonline.com/applications/default/store/product_specific_details.asp?product_category_id=1000012&product_id=1000006" +} +{ + "Tag": [ + "number theory unsolved", + "number theory" + ], + "Problem": "Prove that there are no integer solutions to $x^{6}= y^{5}+24$\r\n\r\nI've tried several mods, but none of them really made any progress.... Any ideas/hints?", + "Solution_1": "It's easy to verify that $y^{5}$ can give following residues $\\mod{31}$:\r\n$0, 1, 5, 6, 25, 26, 30$\r\nAnd $x^{6}$ can give residues:\r\n$0, 1, 2, 4, 8, 16$\r\nContradiction $\\mod{31}$.\r\n\r\nBTW - where can I found the MOP problems?", + "Solution_2": "Oh, I got the MOP problems from a friend. They're not online :( \r\n\r\nBTW, how were you able to calculate all those residues so quickly? Is there a faster way than raising 0,1,2...30 to the 5th and 6th power?", + "Solution_3": "Well, if you know that there are exactly 7 residues which fifth powers can give, and exactly 6 residues which sixth powers can give it's quite easy to calculate.", + "Solution_4": "Generally, if you want to restrict the value of $x^{k}$ for a given mod $m$ you want $gcd(k, \\varphi(m))$ to be large. In this case $31$ is $m$ such that $5, 6 | \\varphi(31)$ which is your best bet.\r\n\r\nA friend of mine gave a related problem, which I think was something like\r\n\r\n$x^{4}= y^{3}+1999$\r\n\r\nWhich gave a contradiction $\\bmod 13$.", + "Solution_5": "I have the idea using Camichael function , and it gave me the way to use mod 31. \nAnd after that we prove like TomciO . But I wonder how we can get this \n$y^{5}$ can give following residues $\\mod{31}$:\n$0, 1, 5, 6, 25, 26, 30$ ", + "Solution_6": "[quote=ht2000]But I wonder how we can get this \n$y^{5}$ can give following residues $\\mod{31}$:\n$0, 1, 5, 6, 25, 26, 30$[/quote]\nAs explained above, we know that there are exactly $\\frac{\\varphi(31)}{5}=6$ non-zero residues of fifth powers $\\mod 31$. Since $1,-1$ are such values and the set of such residues is multiplicative, it suffices to consider $2^5 \\equiv 243 \\equiv -5 \\mod 31$ to see that these $6$ residues must be $\\pm 1, \\pm 5, \\pm 5^2$.\n", + "Solution_7": "[quote=Tintarn][quote=ht2000]But I wonder how we can get this \n$y^{5}$ can give following residues $\\mod{31}$:\n$0, 1, 5, 6, 25, 26, 30$[/quote]\nAs explained above, we know that there are exactly $\\frac{\\varphi(31)}{5}=6$ non-zero residues of fifth powers $\\mod 31$. Since $1,-1$ are such values and the set of such residues is multiplicative, it suffices to consider $2^5 \\equiv 243 \\equiv -5 \\mod 31$ to see that these $6$ residues must be $\\pm 1, \\pm 5, \\pm 5^2$.[/quote]\nWould you mind explaining to me more clearly about this \"we know that there are exactly $\\frac{\\varphi(31)}{5}=6$ non-zero residues of fifth powers\". Do we have the general one ??What theorem tells us ? Please help. ", + "Solution_8": "[quote=ht2000]\nWould you mind explaining to me more clearly about this \"we know that there are exactly $\\frac{\\varphi(31)}{5}=6$ non-zero residues of fifth powers\". Do we have the general one ??What theorem tells us ? Please help.[/quote]\nLet's look at the possible residues of $k$-th powers $\\mod p$ for an odd prime $p$. Well, then we famously have a primitive root $\\mod p$ i.e. an element $g \\mod p$ such that $g,g^2,\\dotsc,g^{p-1}$ cover all the non-zero residues $\\mod p$. Then what are the $k$-th powers? Well, these are exactly $g^k, g^{2k},g^{3k},\\dotsc$. So supposing that $k \\mid \\varphi(p)=p-1$, this gives exactly $\\frac{p-1}{k}$ non-zero residues of $k$-th powers $\\mod p$. This is the situation above where $p=31, k=5$.\nHowever, the argument easily gives a more general result, namely that there are exactly $\\frac{p-1}{(k;p-1)}$ non-zero residues of $k$-th powers $\\mod p$.\nThese results can be generalized to prime powers $p^n$ as modulus and also $2^n$ (but then this is more difficult since there is no primitive root modulo $2^n$ for $n \\ge 3$.\n\n" +} +{ + "Tag": [ + "geometry", + "3D geometry", + "rectangle", + "trigonometry", + "trig identities", + "Law of Cosines" + ], + "Problem": "Find the measure of the obtuse angle formed by two diagonals of a cube.", + "Solution_1": "[quote=\"rnwang2\"]Find the measure of the obtuse angle formed by two diagonals of a cube.[/quote]\r\n\r\n[hide]Let the obtuse angle be $x$. WLOG let the side of the cube be 2. The rectangle that contains both diagonals and is within the cube has sides $2$ and $2\\sqrt{2}$. So the side lengths of the triangle containing $x$ are $2\\sqrt{2}, \\sqrt{3}, \\sqrt{3}$. Using law of cosines, $8=3+3-2\\cdot 3\\cdot \\cos x\\Rightarrow x=\\boxed{\\cos^{-1}\\frac{-1}{3}}$.[/hide]" +} +{ + "Tag": [ + "algebra", + "polynomial", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Show that the set $ \\{(x,y,z): z^2\\equal{}x^2\\plus{}y^2\\}$ is connected.", + "Solution_1": "The set of solutions in $ \\mathbb{R}^n$ or $ \\mathbb{C}^n$ to any homogeneous polynomial equation (or system of homogeneous polynomial equations) is path connected- you can take a line to the origin, and then come back out on another line.\r\n\r\nWe do need to be working over a connected topological field.", + "Solution_2": "i dont understand what you're saying.\r\nin this case how do we show this set is connected?", + "Solution_3": "If $ (x,y,z)$ is on the surface, so is $ (ax,ay,az)$. Follow that line in, and then take another line out to your other point $ (x',y',z')$.\r\n\r\nWe get something quite a bit stronger than being connected- it's \"star-shaped\", which makes it simply connected among other things." +} +{ + "Tag": [ + "function", + "algebra unsolved", + "algebra" + ], + "Problem": "Evaluate $ a(4,4)$ for the function $ a(m,n)$, which is defined for integers $ m,n \\ge 0$ by\r\n\\[ \\begin{array}{l}\r\na(0,n) \\equal{} n \\plus{} 1,\\textup{ if } n \\ge 0; \\\\\r\na(m,0) \\equal{} a(m \\minus{}1, 1), \\textup{ if } m \\ge 0; \\\\\r\na(m,n) \\equal{} a(m \\minus{} 1, a(m, n\\minus{}1)), \\textup{ if } m \\ge 0, \\textup{ and } n \\ge 0.\r\n\\end{array}\\]", + "Solution_1": "$ a(1,n)\\equal{}n\\plus{}2,a(2,n)\\equal{}2n\\plus{}3,a(3,n)\\equal{}2^{n\\plus{}3}\\minus{}3$, \\[ a(4,n)\\equal{}2^{2^{...2^{13}}}\\minus{}3.\\] (n times 2 in power)", + "Solution_2": "Thanks, that's the answer I just worked out by hand - no wonder the computer program I wrote can't compute the answer 0.0", + "Solution_3": "This is IMO 1981, Day 2, Problem 6 see \r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=366648#366648[/url]" +} +{ + "Tag": [ + "algebra unsolved", + "algebra" + ], + "Problem": "Prove that the number $ 317^{4012}$ has at least 10031 digits.", + "Solution_1": "[hide]$ 317^{4012} \\equal{} (317^2)^{2006} \\equal{} (100489)^{2006} > (10^5)^{2006} \\equal{} 10^{10030}$ \n\nThus, $ 317^{4012}$ must have at least $ 10031$ digits as desired. [/hide]", + "Solution_2": "Is it true for 10032 digits???", + "Solution_3": "[quote=\"ridgers\"]Is it true for 10032 digits???[/quote]\r\n\r\nSince $ 4012\\log(317)\\equal{}10034,2498...$, the required number has exactly $ 10035$ digits.\r\n\r\nFor the initial problem, Xantos demo seems to me the best and simpliest one and you can also use it for 10032 :\r\n\r\n$ 317^{4012}\\equal{}(100489)^{2006}\\equal{}1.00489^{2006}10^{10030}$ $ >(1\\plus{}2006\\times 0.00489)10^{10030}$ $ \\equal{}10.80934\\times 10^{10030}$ $ >10^{10031}$ and so at least $ 10032$ digits." +} +{ + "Tag": [ + "function", + "complex analysis", + "real analysis", + "real analysis theorems" + ], + "Problem": "If the Laplace transform (one-sided) of a function exists for any s complex\r\ndoes it mean it can be expressed as a product of a constant and of terms (1-s/l_i)\r\nwhere l_i are its roots?\r\nie L(f,s) = k*Prod 1->00 (1-s/l_i) where k = L(f,0), L(f,l_i) = 0 and i belongs in N.\r\nIf not what else is necessary?Do we need it to be analytic and what exactly we need to check?\r\nPlease elaborate on that.\r\n\r\nThnx\r\nNick", + "Solution_1": "The transform is automatically analytic- you can differentiate under the integral. On the other hand, that form is simply inadequate to describe an entire function- you need to multiply by an exponential $e^{f(x)}$.\r\n\r\nA simple example: Suppose that $f(x)=1$ for $0 Those who know calculus should know this.\r\n\r\nWhich one would you prefer integrating?\r\n\r\nThis: $ \\int\\frac{2x^3\\minus{}6x}{\\sqrt{x^4\\minus{}6x^2\\plus{}3}} {\\rm d} x$\r\n\r\nOr this: $ \\int\\frac{2x\\sqrt{x^{10}\\minus{}12x^8\\plus{}48x^6\\minus{}72x^4\\plus{}27x^2}}{|x|(x^4\\minus{}6x^2\\plus{}3)} {\\rm d} x$\r\n\r\n?\r\n\r\n(Yes, they are equivalent. Mostly.)", + "Solution_7": "Because many times the 2nd version is easier to work with.", + "Solution_8": "[quote=\"SMCount\"]Well it is specified to the [i]denominator[/i].[/quote]\r\n\r\nCorrect, but I think harekrishna got the idea that it makes an irrational number rational, which makes no sense.", + "Solution_9": "Rationalizing puts expressions into a [i]normal form[/i], which makes it easier to decide when two expressions are equal. For example, I claim that\r\n\r\n$ \\frac{1 \\minus{} \\sqrt{1 \\minus{} 4x}}{2x}$\r\n\r\nand\r\n\r\n$ \\frac{2x}{1 \\plus{} \\sqrt{1 \\minus{} 4x} }$\r\n\r\nare equal. Is this obvious? \r\n\r\nOne application familiar to those who know a little number theory is writing down algebraic number fields. I claim that the set of numbers\r\n\r\n$ \\{ a \\plus{} b \\sqrt{2} \\plus{} c \\sqrt{3} \\plus{} d \\sqrt{6} | a, b, c, d \\in \\mathbb{Q} \\}$\r\n\r\nand\r\n\r\n$ \\{ \\frac{a \\plus{} b \\sqrt{2} \\plus{} c \\sqrt{3} \\plus{} d \\sqrt{6} }{e \\plus{} f \\sqrt{2} \\plus{} g \\sqrt{3} \\plus{} h \\sqrt{6} } | a, b, c, d, e, f, g, h \\in \\mathbb{Q} \\}$\r\n\r\nare equal. Is this obvious?", + "Solution_10": "There's an old-fashioned reason for it - a reason that dates back to my time. (I was already in college by the time hand calculators became common.) The reason has to do with arithmetic and writing reasonable decimal approximations.\r\n\r\nSuppose that you got an answer of $ \\frac {2}{\\sqrt {3}}$ and you want some kind of reasonable approximations. You do know that $ \\sqrt {3}\\approxeq 1.732,$ (Tables of square roots were available, and we were also taught how to find square roots with pencil and paper. Besides, I have that much of $ \\sqrt {3}$ memorized.) But if you write $ \\frac {2}{1.732},$ you would still need to do some time-consuming long division to finish this off.\r\n\r\nGo ahead and rationalize the denominator: $ \\frac {2\\sqrt {3}}{3}.$ Now, multiplication and division by small integers is much quicker than full-blown long division. We can say $ \\frac {2\\cdot 1.732}{3} \\equal{} \\frac {3.464}{3}\\approxeq 1.155.$\r\n\r\nRationalizing numerators or denominators is a valuable algebraic skill, and you have to be flexible enough to be able and willing to apply it in [i]either[/i] direction. But the bias that says always rationalize the [i]denominator[/i] - I think that's about 50% a holdover from the precalculator era.\r\n\r\nAnd it's precisely because I think that \"always rationalize the denominator\" is in part a holdover from that precalculator era that I say things in class like $ ``\\cos (45^{\\circ}) \\equal{} \\frac1{\\sqrt {2}}\"$ without apologizing.\r\n\r\nBut I do rationalize denominators more often than numerators anyway - and a lot of the reason for that is the justification that t0rajir0u just gave.", + "Solution_11": "This reminds me of the an old AHSME throw-away, which said something along the lines of:\r\nIf you have an equation of the form $ ax^{2} \\plus{} bx \\plus{} c$ then its solutions will be of the form:\r\nAnd they had all of these bogus quadratic equations, except one...:\r\n$ x \\equal{} \\frac {2c}{ \\minus{} b \\pm \\sqrt {b^{2} \\minus{} 4ac}}$.\r\nAnd then there was the answer choice $ \\text{none of these}$ (this is why, of course, it became a throw-away :P )" +} +{ + "Tag": [ + "function", + "absolute value" + ], + "Problem": "Observe this first inequality.\r\n\r\n$ |a| \\geq b$\r\n\r\n$ a \\geq b \\vee-a \\geq b$\r\n\r\n$ a \\geq b \\vee a \\leq-b \\therefore$\r\n\r\n$ \\boxed{|a| \\geq b \\Longrightarrow a \\leq-b \\vee a \\geq b}$\r\n\r\nThe solution attained is correct. However, I apply the same method with $ |a| \\leq b$ and apparently I can go a step further.\r\n\r\n$ |a| \\leq b$ \r\n\r\n$ a \\leq b \\vee-a \\leq b$ \r\n\r\n$ a \\leq b \\vee a \\geq-b \\therefore$ \r\n\r\n$ \\boxed{|a| \\leq b \\Longrightarrow a \\geq-b \\vee a \\leq b}$\r\n\r\nThe solution above can apparently be simplified to the following:\r\n\r\n$ \\boxed{|a| \\leq b \\Longrightarrow-b \\leq a \\leq b}$\r\n\r\nWhy is this so?", + "Solution_1": "[quote=\"Prime factorization\"]Why is this so?[/quote]\r\n\r\nIt is not so.\r\n\r\n$ a \\leq b \\leq c$ is the same as $ a \\leq b \\wedge b \\leq c$. What you have is a $ \\vee$.", + "Solution_2": "It is true that the claimed result does not follow, but the claimed result is actually true, i.e. for real numbers, $ |a| \\leq b$ implies $ -b \\leq a \\leq b$. (In fact, the two are equivalent.) The trick is that in going from $ |a|\\leq b$ to $ a \\leq b \\vee-a \\leq b$, you've actually tossed away some information -- the latter is not equivalent to the former (it's weaker). In order to get the stronger result, you have to use more properties of the absolute value function than just $ |a| = a \\vee |a| =-a$. (Note that there are other functions which share the property $ f(a) = a \\vee f(a) =-a$ which do not share the property $ f(a) \\leq b \\implies-b \\leq a \\leq b$.)", + "Solution_3": "Do you mind showing me what I missed :blush: I really can't figure out which other property of the absolute value function would be applied to get the stronger inequality.", + "Solution_4": "It's not so much a different property as it is a modification of the given property: $ |a| = a \\vee |a| =-a$ doesn't carry any information about [i]when[/i] $ |a| = a$ (thus my comment about other functions which share this property) -- that's what you're going to have to work with. In other words, break it into cases.", + "Solution_5": "Again, not sure if this is right.\r\n\r\nFor the case where $ |a| \\leq b$ we get:\r\n\r\n$ (a \\leq b \\vee-a \\geq-b) \\land (a \\geq-b \\vee-a \\leq b)$\r\n\r\nOr we could simply say $ (1 \\vee 2) \\land (3 \\vee 4)$\r\n\r\nOur possibilities are\r\n\r\n$ 1 \\land 3$\r\n$ 1 \\land 4$\r\n$ 2 \\land 3$\r\n$ 2 \\land 4$\r\n\r\nAll possible scenarios prove to be equivalent to $ a \\leq b \\land a \\geq-b$ which of course means the same thing as $ -b \\leq a \\leq b$.\r\n\r\n\r\n\r\nFor the case where $ |a| \\geq b$ we get:\r\n\r\n$ (a \\geq b \\vee-a \\leq-b) \\land (a \\leq-b \\vee-a \\geq b)$\r\n\r\nOr we could simply say $ (1 \\vee 2) \\land (3 \\vee 4)$\r\n\r\nAgain our possibilities are\r\n\r\n$ 1 \\land 3$\r\n$ 1 \\land 4$\r\n$ 2 \\land 3$\r\n$ 2 \\land 4$\r\n\r\nSince $ |a| \\geq 0$, all possible scenarios are wrong. I suppose this means that for this case I could then finish with the approach I took in my first post... ??? :huh:", + "Solution_6": "Two problems: the first is structural. Changing $ a \\leq b$ to $ a \\leq b \\wedge-a \\geq-b$ doesn't add new information: the two are equivalent. Since [i]by the time you've reached your first step[/i] in the original derivation you've lost equivalence with what you're trying to prove here, taking something equivalent to the first step cannot be strong enough.\r\nNow, I note that you don't actually have what I expected you to have when I wrote the above sentence: you have a $ \\vee$ on the inside and a $ \\wedge$ between the two terms. This suggests to me that you've simply switched your $ \\vee$s for $ \\wedge$s and vice-versa.\r\n\r\nHere's a stronger hint: $ F \\Longleftrightarrow (a > 0 \\wedge F) \\vee (a < 0 \\wedge F) \\vee (a = 0 \\wedge F)$ for any formula $ F$. (This is the principle of trichotomy. In the case of absolute values, one often collapses the third term with one of the other two.)", + "Solution_7": "I'm really sorry, nothing's working for me. Can you please post the solution JBL?", + "Solution_8": "are we trying to prove that $ |a|0$, so $ |b|=b\\implies |b|>|a|$ and $ |a|\\ge a$, so the claim follows.", + "Solution_9": "Thank you Altheman, I just found the solution and am 100% sure it's correct!\r\n\r\n$ |a| \\leq b$ and $ |a| \\geq 0$ then as you said, b is also greater than 0. Thus it follows that making $ b < 0$ would make $ a$ greater. All that is required is a sign change, thus we get our desired result: $ -b \\leq a \\leq b$.\r\n\r\nI suppose we could then make the following assertion for any general case (this might be wrong) $ \\forall a,b \\in \\mathbf{R}$ we have $ a \\leq b \\land a \\geq 0 \\Longrightarrow a \\geq-b$\r\n\r\nFor $ |a| \\geq b$, since $ |a| \\geq 0$ we have $ b \\in\\mathbf{R}\\leq a$ and thus the method in my first post would suffice.", + "Solution_10": "Could you explain what you are trying to do? My proof seems to be perfectly fine. Is there a particular reason why you are forcing yourself into a particular proof structure?\r\n\r\nI think your 'or' operations are an ugly way to approach this. You are loosing information.\r\n\r\n$ |a|$ means: $ a$ if $ a\\ge 0$, $ -a$ if $ a<0$.\r\n\r\nYour 'or' operation looses information because it merely states when the two results, but not when they occur.", + "Solution_11": "In your proof, you said $ |a| < b \\Longrightarrow a < |b|$ but I don't see how this implies that $ -b \\leq a \\leq b$. Perhaps I wasn't clear on the new approach I used.\r\n\r\n\r\nGiven $ |a| \\leq b$\r\n\r\n$ |a| \\geq 0 \\Longrightarrow b \\geq 0 \\Longrightarrow-b \\leq 0 \\therefore |a| \\geq-b$\r\n\r\nIt follows that $ -b \\leq |a| \\leq b$ \r\n\r\nNow, regardless of whether $ a \\geq 0$ or $ a \\leq 0$, we have\r\n\r\n$ -b \\leq a \\leq b$\r\n\r\nPerhaps your problem is with proving $ |a| \\geq b \\Longrightarrow a \\leq-b \\vee a \\geq b$ with the method I used in the first post, which is understandable. I'll try and find a better method as soon as I can." +} +{ + "Tag": [ + "conics", + "parabola", + "search", + "function", + "Asymptote" + ], + "Problem": "How do you do this?\r\n\r\nIs there some sort of special code you have to type in?\r\n\r\nAnd also, how do you draw the X and Y axes?", + "Solution_1": "Read the documentation :harhar: \r\n[url]http://asymptote.sourceforge.net/doc/graph.html#graph[/url]", + "Solution_2": "lol M. Ivaldi :lol: (I like how you used the emoticon)\r\n\r\nhmm there's also [url=http://www.artofproblemsolving.com/Forum/search.php]the search function[/url]. There already have been several posts regarding this. :)" +} +{ + "Tag": [ + "inequalities proposed", + "inequalities" + ], + "Problem": "Let $a,b,c$ be three length-sides of a triangle such that $ab+bc+ca=1$. Prove that\r\n\\[\\frac{1+a^{2}}{a^{3}+b^{3}}+\\frac{1+b^{2}}{b^{3}+c^{3}}+\\frac{1+c^{2}}{c^{3}+a^{3}}\\ge \\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}+\\frac{27abc}{a^{2}+b^{2}+c^{2}}. \\]", + "Solution_1": "Is it too hard? My solution is in 6 line." +} +{ + "Tag": [], + "Problem": "Hey all you people out there that are interested in baseball listen up! Has any one been watching the little league world series? If not your missing out cause it is [size=150]alsome![/size]. Also if anyone has any friends in it let me know who they are and what team theyare on. If there is any one on here from Richmond,Texas give a holla for our Team! To tell the truth I hope that Richmond wins it!\r\n\r\n\r\n\r\n\r\n\r\n\r\nGmann!", + "Solution_1": "Ah! I forgot about the Little League World Series this year! Oh well, it's kind of lost it's touch these last years, probably because I've just grown out of it.", + "Solution_2": "Yeah Well Panoma beat california In a grulling defeat to win the title. I thought that California whould do better than they did. Oh well I guess it can't be helped, but good job to Panoma for the win.", + "Solution_3": "Is the little league world series just for the U.S., for for the whole world?", + "Solution_4": "Yes, it's a world-wide competition." +} +{ + "Tag": [ + "projective geometry", + "geometry unsolved", + "geometry" + ], + "Problem": "Given a line $ r$, a point $ P$ $ \\notin r$ and a circumference (with its center in the plane). Show how to construct a line $ s$ such that $ P$ $ \\in s$ and $ s\\parallel\\ r$, using only a ruler without any marks/scale.", + "Solution_1": "We recall [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25036]this[/url] theorem\r\nWe will also use the theorem \"If point A lies on the polar of point B, then point B lies on the polar of A\" [url=http://www.cut-the-knot.org/pythagoras/LaHire.shtml#Explanation](La Hire)[/url]\r\n\r\nSuppose that we've constructed the line $ s$. Let $ R,S$ be the pole of $ r,s$ respectively. Then $ OR\\perp r$ and $ OS\\perp s$, so $ O,R,S$ are collinear\r\nSo we need to construct the pole of line $ r$ \r\n\r\n[b]\"ruler only\" construction of polar line[/b]\r\nLet $ F$ be a point not on a given circle $ m$\r\nWe construct two different lines $ l_1,l_2$ passing through $ F$\r\nLet $ A,B \\in l_1 \\cap m$ and $ C,D \\in l_2 \\cap m$\r\nAccording to the above first theorem, the points $ AD\\cap BC$ and $ AC\\cap BD$ define the polar line of $ F$ with respect to $ m$\r\n\r\n\r\nWe choose 2 distinct points on the line $ r$ and we bring their polar lines. \u039fbviously, the intersection point of the 2 polar lines is the pole $ R$ of the line $ r$\r\n\r\nWe construct the polar line $ p$ of the point $ P$. Let $ S\\in p\\cap OR$\r\n\r\n$ S$ lies on the polar of $ A$, hence $ A$ lies on the polar of $ S$\r\n\r\nSo, we finally construct the polar line $ s$ of the point $ S$\r\n\r\nAlso, $ s$ is perpendicular to $ OS$, so $ s\\parallel r$ and passes through $ P$" +} +{ + "Tag": [ + "algebra", + "polynomial", + "algebra proposed" + ], + "Problem": "Find all polynomials $ P: R\\minus{}>R$ such as:\r\n$ P(1)\\equal{}1$\r\n$ x^2P(x\\minus{}1)P(x\\plus{}1) \\equal{} P^2(x)(x\\minus{}1)(x\\plus{}1)$", + "Solution_1": "[quote=\"matex (L)(L)(L)\"]Find all polynomials $ P: R \\minus{} > R$ such as:\n$ P(1) \\equal{} 1$\n$ x^2P(x \\minus{} 1)P(x \\plus{} 1) \\equal{} P^2(x)(x \\minus{} 1)(x \\plus{} 1)$[/quote]\r\nSetting $ f(x) \\equal{} ln \\left ( \\frac {P(x)}{x} \\right )$ we get $ f(x \\minus{} 1) \\plus{} f(x \\plus{} 1) \\equal{} 2f(x)$ which is just Jensen's equation. Having the only solutions $ f(x) \\equal{} ax \\plus{} b$ Thus we have $ P(x) \\equal{} x \\cdot e^{ax \\plus{} b}$ Giving $ a \\equal{} 0$ since $ P(x)$ is a polynomial. and thus the only solutions are $ P(x) \\equal{} ax$. (Of course i've assumed that $ x,P(x) \\neq 0$, but you get the main idea :) )", + "Solution_2": "What suggested to You putting something like this $ f(x) \\equal{} ln \\left ( \\frac {P(x)}{x} \\right )$?\r\nDid You guess it?", + "Solution_3": "I think I found another solution.Correct me if I'm wrong. x=0 => P(0)=0, hence P(x)=xQ(x), and using that we get\r\n$ x^2(x\\minus{}1)(x\\plus{}1)Q(x\\minus{}1)Q(x\\plus{}1)\\equal{}xQ(xQ(x))(x\\minus{}1)(x\\plus{}1)$ assume $ x, x\\minus{}1, x\\plus{}1 \\neq 0$. We can write $ xQ(x\\minus{}1)Q(x\\plus{}1)\\equal{}Q(xQ(x))$, if we assume deqQ=m>0 or m<0 => $ deg LHS\\equal{}2m\\plus{}1$ and $ deg RHS\\equal{}m^2\\plus{}m$ which is a contradiction (odd=even). Hence degQ=0 => Q(x)=a but P(1)=1 implies Q(x)=1 => P(x)=x.", + "Solution_4": "[quote=\"matex (L)(L)(L)\"]What suggested to You putting something like this $ f(x) \\equal{} ln \\left ( \\frac {P(x)}{x} \\right )$?\nDid You guess it?[/quote]\r\nI quickly realised that it was equivalent to:\r\n$ \\frac{P(X\\minus{}1)}{X\\minus{}1} \\cdot \\frac{P(X\\plus{}1)}{X\\plus{}1} \\equal{} \\left ( \\frac{P(X)}{X} \\right ) ^2$. So the first thing i did was putting $ g(x) \\equal{} \\frac{P(X)}{X}$ where $ g(x): \\mathbb{R} \\backslash \\{\\minus{}1,0,1\\} \\to \\mathbb{R}$.\r\nSo we would have $ g(x\\minus{}1) \\cdot g(x\\plus{}1) \\equal{} g(x)^2$. Then I just thought i would be nice if i'd say $ f(x) \\equal{} ln(g(x))$ cause in that way we would add instead of multiply. But then we should remember to define $ f(x): \\mathbb{R} \\backslash \\{\\minus{}1,0,1\\} \\to \\mathbb{C}$! And we also get some problems if $ g(x) \\equal{} 0$.. \r\n\r\nAnyways, limes123's solution is probably better ;)" +} +{ + "Tag": [ + "MATHCOUNTS", + "email" + ], + "Problem": "i heard that for $ \\$$300 some dudes will give u 1000000+ practice problems, private training, etc for MC. it was sent to my email inbox. i think it is a hoax or something.\r\n\r\nWAT DO U GUYS THINK?\r\n\r\ni will try to upload the link soon...", + "Solution_1": "Ya...most likely its a hoax. They just want your money.", + "Solution_2": "If something sounds too good to be true, it's most likely to be fake..\r\nI once received a mail saying if i call this TOLL FREE number, i'll be signed up in a million dollar giveaway..\r\nI've got suspicious so I wrote the same addressee name down on Google :D and found out that other people got the same thing. once they called, they were constantly spammed and annoyed with phone calls saying to buy magazines :rotfl: :rotfl: \r\n\r\nso basically, unless you investigate it thoroughly, it's most likely to be FALSE" +} +{ + "Tag": [ + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "A set $T$ is called naughty if for any two (not necessarily distinct) elements $u,v$ of $T$, $u+v$ does not belong to $T$. Prove that\r\n\r\n(a) a naughty subset of $S = \\{1,2,...,2006\\}$ has at most $1003$ elements\r\n\r\n(b) every set $S$ of $2006$ positive numbers contains a naughty subset having $669$ elements.", + "Solution_1": "(a) is easy suppose the biggest number in naughty subset A is n. Then if a is in A then n-a is not in A. Therefore only half of the numbers less then n are in A, so $|A|\\leq 1003$", + "Solution_2": "unless im missing something, suppose $n$ is the largest element in $S$. Any subset of $S$ containing $n$ is naughty, because it would require $n+n$ to be in the subset which is not possible. So now just pick 668 more integers. More generally, if a set has a finite number of elements $k$, there is a naughty subset of order $a$ for all $1\\lte a\\lte k$", + "Solution_3": "[quote]Any subset of S containing n is naughty,[/quote]\r\n[b]gopherhole112[/b], it is not true because S may contain a and b such that a+b=n.", + "Solution_4": "It was posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=77010 ." +} +{ + "Tag": [], + "Problem": "Ticked-Off Todd started off in Tokyo at 10:12 AM local time. He walked continously until approximately 7:32 PM local time. If his pace was steady at 740 meters every 16 minutes, \r\n\r\n- How many miles did Ticked-Off Todd walk in Tokyo?\r\n- What was his average mph?", + "Solution_1": "So should we convert miles to meters?\r\n\r\nBecause I have 25900 meters\r\n\r\nor 46.25 meters per min or 2775 mph", + "Solution_2": "TOT walked for 9 hr 20 min, or 560 min. 560 * 740/16 = 25900 m.\r\n\r\n(Just expanding on Urc's first part. the second is a simple conversion, as the rate is already given, and does not need the other info. :wink: )", + "Solution_3": "there is a rational conversion between meters and miles but it is uber...\r\nassuming 25900 meters is right...\r\n25900 meters = 2590000 centimeters = $ \\frac{2590000}{2.54} \\equal{} \\frac{129500000}{127}$ inches = $ \\frac{(\\frac{129500000}{127})}{12} \\equal{} \\frac{32375000}{381}$ feet = $ \\frac{(\\frac{32375000}{381})}{5280} \\equal{} \\frac{809375}{50292} \\sim 16.0935$ miles \r\nphew\r\n\r\nnow we convert 740 meters per 16 minutes into miles per hour\r\n$ \\frac{740 m}{16min} \\equal{} \\frac{740 \\cdot 60 m}{16 hr} \\equal{} 185 \\cdot 15 \\cdot \\frac{meter}{hour}\\equal{} 277500 \\cdot \\frac{cm}{hr} \\equal{} \\frac{277500}{2.54} \\cdot \\frac{in}{hr} \\equal{} \\frac{13875000}{127} \\cdot \\frac{in}{hr}$\r\nthere are 12 inches in a foot and 5280 feet in a mile so there are 63360 inches in a mile, so\r\n$ \\frac{13875000}{127} \\cdot \\frac{in}{hr} \\equal{} \\frac{(\\frac{13875000}{127})}{63360} \\frac{mi}{hr} \\equal{} \\frac{346875}{201168} mph \\sim 1.7243 mph$ \r\nphew\r\n\r\nfor quick reference, 1 mile = 1.609344 km" +} +{ + "Tag": [ + "geometry" + ], + "Problem": "A right triangle is inscribed in a circle with a diameter $ 100$ units long. What is the maximum area of the triangle, in square units?", + "Solution_1": "The law is that the right triangle inscribed in a circle means that diameter=hyptonuse.\r\nI guess that it is a 3-4-5 triangle and that 60*80/2=2400\r\nCan someone explain please?", + "Solution_2": "I think that for maximum area, you would need a right isosceles triangle. Don't ask me why... but it's true...\r\n\r\nThe leg would be $ \\frac{100}{\\sqrt2}$, or $ 50\\sqrt{2}$.\r\n\r\nThus, $ \\frac{(50\\sqrt{2})^2}{2}$ gives the answer of $ \\boxed{2500}$." +} +{ + "Tag": [ + "inequalities", + "triangle inequality", + "calculus", + "calculus computations" + ], + "Problem": "i came across this rather peculiar characteristic of conditionally converging series in my book:\r\n\r\nit is said that for any conditionally converging alternating series, we can always rearrange the terms so that the sum of the series changes or become divergence.\r\n\r\ne.g:\r\n\r\n$ln2 = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + ......$\r\n\r\nif we arrange it so that:\r\n$(1 - \\frac{1}{2} - \\frac{1}{4}) + (\\frac{1}{3} - \\frac{1}{6} -\\frac{1}{8}) + (\\frac{1}{5} - \\frac{1}{10} - \\frac{1}{12}) + .....)$\r\nthen we add the first two terms in each group, and factor out the $\\frac{1}{2}$, we will get:\r\n\r\n$\\frac{1}{2}(1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + ......)$,\r\nwhich is $\\frac{1}{2}ln2$\r\n\r\ncan someone explain how that happens? since we're taking the sum of the same series, why is the sum different if we rearrange the sequence?\r\nis it because we \"pulled\" the negative terms in front?", + "Solution_1": "[quote=\"violetcraze\"]can someone explain how that happens? since we're taking the sum of the same series, why is the sum different if we rearrange the sequence?\nis it because we \"pulled\" the negative terms in front?[/quote]\r\n\r\nHeh, I just taught a class about this on Wednesday. Here's an (informal) way to make any conditionally convergent series converge to any real value. If it's conditionally convergent, it can be split into two divergent sub-sequences, one of positive terms and the other of negative terms. Say you have a target value of $a \\geq 0$. Take positive terms until you're larger than $a$ for the first time, then take negative terms until you're smaller than $a$ again, then take positive terms until you're larger than $a$, and repeat. Because the series converges, the terms eventually get to 0, so we eventually are always very close to $a$, so this re-arrangement of the series converges to $a$. I think this displays the basic mechanism of what's going on quite well, but I can give more details if you want.", + "Solution_2": "[quote=\"JBL\"] I think this displays the basic mechanism of what's going on quite well, but I can give more details if you want.[/quote]\r\n\r\nwow, thanks a lot. :D we are all wondering why and even my lecturer can't explain that. \r\np/s: i would like more details if possible, please :)", + "Solution_3": "Your intution about \"pulling\" the negative values forward is exactly what's happening. You can also make the sequence diverge in either direction or oscillate, in the same way that JBL suggested. Just take way more negative terms than positive ones, but do it in such a way that \"eventually\" you'll still have all the same terms -- then your sequence will go to $-\\infty$. When I say \"eventually\" I don't mean the first $k$ terms of your rearrangement will have the same terms as the first $k$ of the original series -- in that case, you'll get teh same limit. I just mean that when you rearrange, you can only \"push back\" and \"pull forward\"; you can't throw stuff out. \r\n\r\nThis stuff was all thought of and proved by Riemann. :D", + "Solution_4": "The basic distinction here has to do with order: for some reason, the terms of a conditionally convergent series aren't enough to determine it's behavior. Why is this? Well, absolute convergence essentially says that no matter what subsequence we take, it will always converge. So the basic behavior (convergence) of an absolutely convergent sequence is determined directly by the (unordered) terms of the sequence. For a conditionally convergent sequence, this isn't true -- the basic behavior is unclear if I just give you an unordered collection of the terms of the series because it has many divergent sub-series. So that's why order is important: otherwise, there isn't any way to figure out what's going on in conditionally convergent series. Once you realize that, it's not at all suprising that you can abuse the series in various ways. (One thing what I said doesn't show: I haven't given any convincing explanation of why re-ordering an absolutely convergent series always results in the same sum. In fact, I don't know the proof of this fact off the top of my head. Perhaps it is that property which should be considered unusual.)", + "Solution_5": "Why reordering an absolutely convergent series is okay: \r\n\r\nWe may as well assume that all of the terms are nonnegative*. If the series is convergent, then we know that for any $\\epsilon > 0$, the sum of all the terms beyond $N(\\epsilon)$ is less than $\\epsilon$. Rearrange the sequence any way you like. Pick some $\\epsilon$, and get from it the corresponding $N$ from the original sequence. Those first $N$ terms are finite in number, so there exists some $M(N)$ beyond which none of those terms appear in the rearrangement. So the sum of the terms beyond $M$ in the rearrangement has to be less than or equal to the sum of the terms beyond $N$ in the original series (we may have lost some positive terms, but no negative ones -- the critical part of the proof), which is itself less than $\\epsilon$, so the rearranged series converges as well. \r\n\r\n*Triangle inequality:\r\n$|\\sum_{k=0}^n a_k| \\le \\sum_{k=0}^n |a_k|$.\r\nIf $\\sum_k a_k$ is an absolutely convergent series, then both sides of that converge as we take a limit in $n$, and the inequality will hold all the way.\r\n\r\nEdit: Well, I just realized I gave an analytical proof, but no intuitive explanation. Basically, the point is that in a conditionally convergent series, we converge because the negative and positive terms balance and mostly cancel each other out. However, in an absolutely convergent series, we are told that even if all of the negative terms were made positive, it would still converge. So that's saying that we don't need the negative terms to balance the positive ones. Then what we're saying is that in some sense, only finitely many of the terms are \"too big.\" The rest of them don't matter. So no matter how you rearrange the terms, eventually you will get past all the ones that are \"too big\" and what's left doesn't matter. So the series always converges (and to the same limit), no matter how you rearrange it.", + "Solution_6": "But why does it always converge to the same sum?", + "Solution_7": "How about this?\r\nLet $a_n^+=\\begin{cases}a_n& a_n>0\\\\0& a_n\\le0\\end{cases}$ and $a_n^-=\\begin{cases}a_n& a_n<0\\\\0& a_n\\ge0\\end{cases}$. If $\\sum a_n$ converges absolutely, both $\\sum a_n^+$ and $\\sum a_n^-$ converge by comparison; let $\\sum a_n^+=A$ and $\\sum a_n^-=-B$.\r\nGiven $\\epsilon>0$, we can find $N_1$ and $N_2$ such that $A-\\sum_{n=1}^{N_1}a_n^+<\\epsilon$ and $\\sum_{n=1}^{N_1}a_n^- -B<\\epsilon$. Let $a_{n(k)}$ be a rearrangement. We can find $N$ such that $n(k)>N_1$ and $n(k)>N_2$ whenever $k>N$. Then $A\\ge\\sum_{k=1}^{m}a_{n(k)}^+\\ge\\sum_{n=1}^{N_1}a_n^+$ and $\\sum_{n=1}^{N_1}a_n^-\\ge\\sum_{k=1}^{m}a_{n(k)}^-\\ge -B$ whenever $m\\ge n$, so $A>\\sum_{k=1}^{m}a_{n(k)}>A-\\epsilon$ and $-B+\\epsilon>\\sum_{k=1}^{m}a_{n(k)}^\\to -B$. Adding the two inequalities, $A-B+\\epsilon>\\sum_{k=1}^{m}a_{n(k)}>A-B-\\epsilon$ whenever $m>N$, and the rearranged series converges to $A-B$.", + "Solution_8": "Well, again let's look at the first series. Suppose it converges to $C$. We pick $\\epsilon > 0$, and that gives us some $N$. The sum of all the terms after $N$ is less than $\\epsilon$. When we rearrange the sequence, all the first $N$ terms from the original sequence are within the first $M$ terms from the other sequence. The sum of those is within $\\epsilon$ of $C$. The other $M - N$ terms are from the tail of the original sequence, which has sum $< \\epsilon$, so is bounded in sum by $\\epsilon$. As noted before, the tail of the rearranged sequence has sum less than $\\epsilon$, so we can bound the whole thing to within $3\\epsilon$ of $C$ (and maybe closer, I think I'm being over-generous). \r\n\r\nThe critical part is that I'm looking at a sequence of positive terms. Otherwise my claims about the tail being small and the extra $M-N$ terms in the head being small in the rearranged sequence wouldn't be assured. \r\n\r\nheh jmerry got there first, and formalized my claim about being able to assume it's positive, too." +} +{ + "Tag": [], + "Problem": "Explain the differences between how solvents affect that rate of $ S_N1$ and $ S_N2$ reactions.\r\n\r\nAlso, what are the effects of $ \\text{polar protic}$, $ \\text{polar aprotic}$ and $ \\text{non\\minus{}polar aprotic}$ solvents on the rate of nucleophilic reactions.", + "Solution_1": "I think it would be more instructive if you post first your own thoughts.", + "Solution_2": "Actually, my question should be extended a bit. How can one determine whether $ S_N1,S_N2,E1,E2,E1cB$ will occur based on the substrate species, solvent used etc.\r\n\r\nI can understand the part about using $ \\text{polar protic}$ and $ \\text{polar aprotic}$ solvents but I can't think what might happen if $ \\text{non\\minus{}polar aprotic}$ solvents are used because how are they supposed to stabilize the substrate or transition state?" +} +{ + "Tag": [ + "number theory unsolved", + "number theory" + ], + "Problem": "I have a trouble with this problem :\r\n n is a given positive integer. Let M = {x/x=p1^s1 * p2^s2*....*pn^sn}, p1,...,pn are given primes, s1,....,sn are positive integers.\r\n Find the minimum number m such that for every subset M consisting of\r\nm elements, there exists x1,x2,x3,x4 of M satisfying x1*x2*x3*x4=a^4, a is an integer.\r\n I can prove that if m= 3*2^n+1 then it's right, but I wonder if m is mininum or not. Please help me.", + "Solution_1": "Its from IMO 1984 but created by treegoner? :D", + "Solution_2": "I just made a mistake. :D \r\n Perhaps that problem appeared in IMO 1985 not 1984. :)", + "Solution_3": "I created the same problem before you treegoner :D it's somewhere on the forum..." +} +{ + "Tag": [ + "email", + "Columbia" + ], + "Problem": "I got this email from Justin Case [webmaster.adminmail@gmail.com] yesterday and I am very confused. It this a hoax or is real? I don't possibly believe this is real because I have never sent emails to and from Columbia. \r\n\r\n\r\n*****,\r\nRecently as we have observed your account, we have come across emails to and from Columbia speaking of purchasing illegal drugs. These communications with Columbia are illegal because of anti-drug trafficking laws in America. We will be forced to terminate your account and permanently delete you from the system by law. We also will copy your IP address and send the police after you after a week if this has not stopped. Your computer will be confiscated, and you will be jailed. You have a week to stop these activities and communications or these things will happen to you. Now this is a WARNING.\r\nDO NOT SCREW UP!!!!!\r\n\r\n--\r\nJustin Case\r\nmail administrator", + "Solution_1": "I doubt that is real at all. I checked some real gmail admistraters emails. That should be a phony address. You should contact the staff a notifie them about this.", + "Solution_2": "It sounds fake. I doubt professional e-mails would include the words \"screw up\".", + "Solution_3": "That e-mail seems really weird. The \"administrator\" sounds very suspicious. \r\n(Just-In-Case)", + "Solution_4": "[quote=\"plokoon51\"]That e-mail seems really weird. The \"administrator\" sounds very suspicious. \n(Just-In-Case)[/quote]\r\n\r\nNot necessarily. I have a cousin by that name (really, I do) :D \r\n\r\n\r\n\r\nIt's probably fake.", + "Solution_5": "just out of curiosity, did you make the hoax?", + "Solution_6": "[quote=\"MysticTerminator\"]just out of curiosity, did you make the hoax?[/quote]\r\nNo.", + "Solution_7": "um. I meant churchilljrhigh. sorry", + "Solution_8": "Hello everyone,\r\n\r\nHonestly I doubt that the mail is serius. Gmail is a seruis and big company to tell you DON'T SCREW IT!!! Any way, since gmail was created, there are people complaining respect to some aspects fo privacy. An example\r\n\r\nhttp://www.google-watch.org/gmail.html\r\n\r\nI do have, not one but two gmail accounts, and I use them principally when I want to have some file uploaded to a \"private\" section of the web, to be able to access it form anywhere, and also to use it for mailing list, given that, any way, everything that is discussed in such lists is pulbic! \r\n\r\nEither way, if you didn't create the hoax, just contact the real administrator!!!\r\n\r\nBest regards,", + "Solution_9": "[quote=\"MysticTerminator\"]um. I meant churchilljrhigh. sorry[/quote]\r\n\r\nThanks for the enlightenment. Of course, I didn't. Truth be told, the email scared me because it was the first time I have seen \"Personalized Spam\" sent to me. And so deducting from all this, probably one of my friends e-mailed it.", + "Solution_10": "Shouldn't the country's name be spelled ColOmbia?", + "Solution_11": "Maybe she is trafficking drugs =o =o =o\r\n\r\nI'm just kidding :P ,", + "Solution_12": "that's really mean and scary", + "Solution_13": "Well, \r\n\r\nThere is also District of Columbia, and Columbia, New England. On the other hand there is also Columbia House. But the name of the country is really Colombia.\r\n\r\n[quote=\"Treething\"]Maybe she is trafficking drugs =o =o =o\nI'm just kidding :P ,[/quote]\r\n\r\nYou may be kidding, but it should be an acusation that you shouldn't do, not even kidding!\r\n\r\nBest," +} +{ + "Tag": [ + "search", + "MATHCOUNTS" + ], + "Problem": "Here's how this works:\r\n\r\nLook at the [b]last post[/b] of this topic. If it is an image, post a guess as to a word or phrase (<=3 words) that will get that image in Google Image Search (http://images.google.com). If it is a word or phrase, search it in Google Image Search and post the picture.\r\n\r\nInstructions for posting images: When you find the image on GIS, click on it, then click \"view full size image\". Copy the address in your address bar and then post\r\n[code][img][URL OF IMAGE][/img][/code]\r\n\r\nAdditional rules:\r\nTry retrieving images from the first couple pages of the results.\r\nDo not post inappropriate images\r\nDo not post phrases that would usually lead to inappropriate images (Google's filter is good but not perfect)\r\n[color=darkblue]Moderator Notice: Please (for the last time) place all images under hide tags.[/color]\r\n\r\nI'll start\r\n\r\n[hide]\n[img]http://www.teachspin.com/instruments/faraday/farradayrotation_big.jpg[/img]\n[/hide][", + "Solution_1": "Faraday rotation. I have seen it before.\r\n\r\n[hide]\n[img]http://www.fbs.leeds.ac.uk/garden/photos/Tomato-Hydroponics.jpg[/img]\n[/hide]\r\n\r\nCouldn't you just right-click the picture and see the url and go from there?", + "Solution_2": "[hide]\n[img]http://www.mariowiki.com/images/b/b3/Piranha_Plant.JPG[/img]\n[/hide]\r\n\r\nLOL", + "Solution_3": "Piranha Plant.\r\n\r\n[hide]\n[img]http://www.cscc.edu/docs/math/images/ARML.jpg[/img]\n[/hide]\r\n\r\nOk... If you can't get that then.... uh... I would really question your \r\n\r\nsanity...", + "Solution_4": "ARML... DUH!!!\r\n\r\n[hide]\n[img]http://oc.uan.edu.co/obm/obmLogo.gif[/img]\n[/hide]", + "Solution_5": "Olimpiadas Bolivarianas\r\n\r\nNew thing:\r\n\r\n[hide]\n[img]http://www.whitehouse.gov/news/releases/2006/05/images/20060515-1_p051506pm-0142-515h.jpg[/img]\n[/hide]", + "Solution_6": "MATHCOUNTS national competition\r\n\r\n[hide=\"clicky!\"][img]http://www.redhat.com/magazine/025nov06/features/xkcd/stringtheory.jpg[/img][/hide]", + "Solution_7": "redhat magazine\r\n\r\n[img]http://i.a.cnn.net/si/2008/writers/don_banks/02/03/superbowl.snaps/T1_0203_plaxico.jpg[/img]", + "Solution_8": "PLAXICO BURRESS AFTER CATCHING THE SUPER BOWL XLII GAME WINNING TD-PASS!\r\n[hide][img]http://www.cut-the-knot.org/gifs/AoPS.gif[/img][/hide]\r\n\r\nif you can't get this, you dont belong here...", + "Solution_9": "Guys mine hasn't been figured out.", + "Solution_10": "[quote=\"BOGTRO\"]Olimpiadas Bolivarianas\n\nNew thing:\n\n[hide]\n[img]http://www.whitehouse.gov/news/releases/2006/05/images/20060515-1_p051506pm-0142-515h.jpg[/img]\n[/hide][/quote]\r\nLol...I'm in that picture.", + "Solution_11": "[quote=\"hunter34\"]Guys mine hasn't been figured out.[/quote]\r\n\r\nit's tomato hydroponics (i hope...)\r\n\r\n@mz94 = AOPS logo. YAY\r\n\r\n[hide=\"a picture\"][img]http://www.maa.org/editorial/mathgames/PrimeSpiral41.gif[/img][/hide]", + "Solution_12": "[quote=\"hunter34\"]Faraday rotation. I have seen it before.\n\n[hide]\n[img]http://www.fbs.leeds.ac.uk/garden/photos/Tomato-Hydroponics.jpg[/img]\n[/hide]\n\nCouldn't you just right-click the picture and see the url and go from there?[/quote]\n\nTomato hydroponics. (thanks for the idea)\n\n\n\nEDIT: Image location won't help you with this one.\n\n[hide]\n[img]http://www.craftster.org/forum/avataruploads/avatar_2152.jpg[/img]\n[/hide]\r\n\r\n\r\nEDIT 2: To the above: Prime spiral", + "Solution_13": "lisa simpson scream\r\n\r\nIt appears to be cropped.\r\n\r\n[hide][img]http://www.bmumford.com/photo/ballistics/bulb.jpg[/img][/hide]", + "Solution_14": "lightbulb being pierced by pellet gun\r\n\r\n[hide=\"pic\"][img]http://www.filebuzz.com/software_screenshot/full/2661-Image_for_DOS.gif[/img][/hide]", + "Solution_15": "imaging in progress DOS\r\n\r\n[hide=\"picture\"][img]http://msnbcmedia3.msn.com/j/msnbc/Components/Video/060310/nn_savidge_sats_060310.300w.jpg[/img][/hide]", + "Solution_16": "SAT.\r\n\r\n\r\n\r\n[img]http://l.yimg.com/img.tv.yahoo.com/tv/us/img/site/35/30/0000043530_20071001142844.jpg[/img]", + "Solution_17": "[hide][img]http://www.google.com/imgres?q=egg&start=98&num=10&um=1&hl=en&client=firefox-a&rls=org.mozilla:en-US:official&biw=1366&bih=601&tbm=isch&tbnid=_a0b4JoFy6625M:&imgrefurl=http://www.bmumford.com/photo/ballistics/index.html&docid=_jdNLh3GZ3nIpM&w=500&h=520&ei=8i11Ts7oFtHTiALXru2zAg&zoom=1&iact=hc&vpx=165&vpy=118&dur=8408&hovh=229&hovw=220&tx=159&ty=116&sqi=2&page=5&tbnh=119&tbnw=110&ndsp=26&ved=1t:429,r:0,s:98[/img][/hide]" +} +{ + "Tag": [ + "limit" + ], + "Problem": "\u0395\u03c3\u03c4\u03c9 $ F_n$ \u03b7 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03b9\u03b1 Fibonnaci. \u039d\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03b5\u03c4\u03b5 \u03c4\u03bf \u03bf\u03c1\u03b9\u03bf:\r\n\r\n$ \\lim_{n \\rightarrow +\\infty}{\\sqrt[n]{F_n}}$ :)", + "Solution_1": "[hide]$ F_n \\sim \\frac{\\phi^n}{\\sqrt{5}}$, \u03bf\u03c0\u03cc\u03c4\u03b5...[/hide]\r\n\r\nCheerio,\r\n\r\nDurandal 1707", + "Solution_2": "\u03a9\u03c1\u03b1\u03b9\u03b1, \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03b9 \u03bf\u03bc\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b7\u03bc\u03bf\u03c0\u03bf\u03b9\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c4\u03b1\u03c5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1 \u03c4\u03bf\u03c5 binet (\u03c0\u03c9\u03c2?)", + "Solution_3": "\u03a4\u03bf \u03bf\u03c1\u03b9\u03bf \u03c4\u03b7\u03c2 \u03bd-\u03bf\u03c3\u03c4\u03b7\u03c2 \u03c1\u03b9\u03b6\u03b1\u03c2 \u03b9\u03c3\u03bf\u03c5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03bf\u03c1\u03b9\u03bf \u03c4\u03bf\u03c5 \u03bb\u03bf\u03b3\u03bf\u03c5 \u03c4\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf \u03bf\u03bc\u03c9\u03c2 \u03b1\u03c0\u03bf \u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf \u03c4\u03b7\u03c2 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03b9\u03b1\u03c2 fibonacci \r\n\u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c6 \u03b1\u03c6\u03bf\u03c5:\r\n\r\n$ \\frac {f_{n \\plus{} 2}}{f_{n \\plus{} 1}} \\frac {f_{n \\plus{} 1}}{f_n} \\equal{} \\frac {f_{n \\plus{} 1}}{f_n} \\plus{} 1 \\equal{} >$\r\n\r\n$ g_{n \\plus{} 1}g_n \\equal{} g_n \\plus{} 1 \\equal{} >$\r\n\r\n\u03a0\u03b1\u03b9\u03c1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bf\u03c1\u03b9\u03b1\r\n\r\n$ x^2 \\equal{} x \\plus{} 1$\r\n\r\n\u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c1\u03b9\u03b6\u03b1 \u03c4\u03bf \u03c6.", + "Solution_4": "\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5 generating functions... \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03ad\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf generatingfunctionology \u03c4\u03bf\u03c5 Wilf \u03c3\u03b1\u03bd \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 (\u03bd\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c2 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c4\u03b7\u03bd \u03b1\u03c3\u03c5\u03bc\u03c0\u03c4\u03c9\u03c4\u03b9\u03ba\u03ae \u03c3\u03c5\u03bc\u03c0\u03b5\u03c1\u03b9\u03c6\u03bf\u03c1\u03ac \u03c4\u03b7\u03c2 $ F_n$), \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03c9 \u03ba\u03b9 \u03cc\u03c1\u03ba\u03bf...\r\n\r\nCheerio,\r\n\r\nDurandal 1707" +} +{ + "Tag": [ + "integration", + "logarithms", + "calculus", + "function", + "calculus computations" + ], + "Problem": "$ \\int_{2}^{4} \\frac {\\sqrt {ln(9 \\minus{} x)}}{\\sqrt {ln(9 \\minus{} x)} \\plus{} \\sqrt {ln(x \\plus{} 1)}}dx$", + "Solution_1": "I think $ \\sqrt{\\ln (x\\plus{}1)}$ should be $ \\sqrt{\\ln (x\\plus{}3)}$.", + "Solution_2": "No, it's like this in the book, anyway give a hint to the one you know, please.", + "Solution_3": "C'est vrais?\r\n\r\nAnyway, if my opinion would be right, my idea was that since $ \\frac {(9 \\minus{} x) \\plus{} (x \\plus{} 3)}{2} \\equal{} 6$, let $ 6 \\minus{} x \\equal{} t$, we have....", + "Solution_4": "maybe the upper limit of the integral is a 6, and Kunny's idea still works... :D\r\nObviously with $ x \\plus{} t \\equal{} 8$ in this case, of course...\r\n\r\nNote: This is actually a hint, believe it or not...", + "Solution_5": "[hide=\"The Hint\"]Compute:\n\n$ \\int_{a}^{b} \\frac {f(a \\minus{} x)}{f(a \\minus{} x) \\plus{} f(x \\minus{} b)}dx$\nfor every function $ f$\n :D [/hide]", + "Solution_6": "I get it! \r\n\r\n$ \\int_2^4 \\frac {\\sqrt {\\ln(9 \\minus{} x)}}{\\sqrt {\\ln(9 \\minus{} x)} \\plus{} \\sqrt {\\ln(x \\plus{} 1)}}dx \\equal{} \\int_4^6 \\frac {\\sqrt {\\ln(9 \\minus{} x)}}{\\sqrt {\\ln(9 \\minus{} x)} \\plus{} \\sqrt {\\ln(x \\plus{} 1)}}dx$\r\n\r\n$ \\therefore \\int_2^4 \\frac {\\sqrt {\\ln(9 \\minus{} x)}}{\\sqrt {\\ln(9 \\minus{} x)} \\plus{} \\sqrt {\\ln(x \\plus{} 1)}}dx \\equal{} \\frac {1}{4} \\int_2^6 dx \\equal{} 1$." +} +{ + "Tag": [ + "function", + "algebra proposed", + "algebra" + ], + "Problem": "Find all functions $ f: \\mathbb{R}\\rightarrow \\mathbb{R}$ such that for all $x,y\\in\\mathbb{R}$,\n\\[f(xy(x\\plus{}y)) \\equal{} f(x)f(y)(f(x)\\plus{}f(y)).\\]", + "Solution_1": "No ideas?Must I post my solution?", + "Solution_2": "[quote=\"popolux\"]Find all functions $ f: \\mathbb{R}\\rightarrow \\mathbb{R}$ such that $ \\forall x,y\\in\\mathbb{R},f(xy(x \\plus{} y)) \\equal{} f(x)f(y)(f(x) \\plus{} f(y))$[/quote]\r\n\r\nHere is a rather long and not very nice solution. Dont hesitate to post a nicer one :)\r\n\r\nLet $ P(x,y)$ be the assertion $ f(xy(x\\plus{}y))\\equal{}f(x)f(y)(f(x)\\plus{}f(y))$\r\n\r\n1) Constant solutions are $ f(x)\\equal{}0$, or $ f(x)\\equal{}\\frac{1}{\\sqrt 2}$ and $ f(x)\\equal{}\\minus{}\\frac{1}{\\sqrt 2}$\r\nJust solve the equation $ c\\equal{}2c^3$\r\n\r\n2) Non constant solutions are such that $ f(0)\\equal{}0$\r\n$ P(0,0)$ $ \\implies$ $ f(0)\\equal{}2f(0)^3$ $ \\implies$ $ f(0)\\in\\{\\minus{}\\frac{1}{\\sqrt 2},0,\\plus{}\\frac{1}{\\sqrt 2}\\}$\r\n$ P(x,0)$ $ \\implies$ $ f(0)\\equal{}f(x)f(0)(f(x)\\plus{}f(0))$\r\nIf $ f(0)\\neq 0$, we get $ f(0)\\equal{}\\frac{\\epsilon}{\\sqrt 2}$ (where $ \\epsilon\\equal{}\\pm 1$) and $ f(x)^2\\plus{}f(x)f(0)\\minus{}1\\equal{}0$ and so :\r\nEither $ f(x)\\equal{}\\frac{\\epsilon}{\\sqrt 2}$, either $ f(x)\\equal{}\\frac{\\minus{}2\\epsilon}{\\sqrt 2}$\r\n\r\nBut $ P(x,x)$ $ \\implies$ $ f(2x^3)\\equal{}2f(x)^3$. So, $ f(x)\\equal{}\\frac{\\minus{}2\\epsilon}{\\sqrt 2}$ would imply $ f(2x^3)\\equal{}\\frac{\\minus{}8\\epsilon}{\\sqrt 2}$, which is impossible.\r\nSo $ f(x)\\equal{}f(0)$ and $ f(x)$ is constant.\r\nQ.E.D.\r\n\r\n3) In non constant solutions, $ f(a)\\equal{}0$ $ \\iff$ $ a\\equal{}0$\r\nWe already know that $ f(0)\\equal{}0$ (point 2)\r\nIf $ f(a)\\equal{}0$ for some $ a\\neq 0$, then $ P(x,a)$ $ \\implies$ $ f(ax(x\\plus{}a))\\equal{}0$ $ \\forall x$\r\nSince the equation $ ax(x\\plus{}a)\\equal{}b$ has solutions for any $ b$ such that $ a^4\\plus{}4ab\\geq 0$, we get $ f(b)\\equal{}0$ $ \\forall b$ such that $ a^4\\plus{}4ab\\geq 0$\r\nSo $ f(x)\\equal{}0$ $ \\forall x$ with same sign as $ a$.\r\nBut $ b\\equal{}\\minus{}\\frac{a^3}{8}$ is such that $ a^4\\plus{}4ab\\equal{}a^4\\minus{}\\frac{a^4}{2}\\geq 0$ and so $ f(b)\\equal{}0$ with $ b$ opposite sign than $ a$\r\nSo $ f(x)\\equal{}0$ $ \\forall x$ with same sign as $ b$, so with opposite signe as $ a$\r\nQ.E.D.\r\n\r\n4) Non constant solutions are injective\r\nLet $ x\\neq 0$ : $ P(x,\\minus{}x)$ $ \\implies$ $ f(0)\\equal{}f(x)f(\\minus{}x)(f(x)\\plus{}f(\\minus{}x))$ and so $ f(\\minus{}x)\\equal{}\\minus{}f(x)$ (since neither $ f(x)\\equal{}0$, neither $ f(\\minus{}x)\\equal{}0$, according to point 3).\r\nSo, suppose $ f(a)\\equal{}f(b)$, then $ f(\\minus{}b)\\equal{}\\minus{}f(a)$ and $ P(a,\\minus{}b)$ $ \\implies$ $ f(\\minus{}ab(a\\minus{}b))\\equal{}0$ and so $ ab(a\\minus{}b)\\equal{}0$\r\nSo, either $ a\\equal{}b$, either $ a\\equal{}0$ and so $ f(b)\\equal{}0$ and $ b\\equal{}0$, either $ b\\equal{}0$ and so $ f(a)\\equal{}0$ and so $ a\\equal{}0$\r\nQ.E.D.\r\n\r\n5) Non constant solutions are such that $ f(x)\\equal{}x$ $ \\forall x\\in\\mathbb Q$ or $ f(x)\\equal{}\\minus{}x$ $ \\forall x\\in\\mathbb Q$\r\n5.1) $ f(2x)\\equal{}2f(x)$\r\n$ P(x,x)$ $ \\implies$ $ f(2x^3)\\equal{}2f(x)^3$\r\n$ P(2x,\\minus{}x)$ $ \\implies$ $ f(2x(\\minus{}x)(2x\\minus{}x))\\equal{}f(2x)f(\\minus{}x)(f(2x)\\plus{}f(\\minus{}x))$ $ \\implies$ $ \\minus{}f(2x^3)\\equal{}\\minus{}f(2x)f(x)(f(2x)\\minus{}f(x))$\r\n$ \\implies$ $ 2f(x)^3\\equal{}f(2x)f(x)(f(2x)\\minus{}f(x))$\r\n$ \\implies$ $ f(x)(f(2x)\\minus{}2f(x))(f(2x)\\plus{}f(x))\\equal{}0$. So :\r\nEither $ f(x)\\equal{}0$ and so $ x\\equal{}0$ and so $ f(2x)\\equal{}2f(x)\\equal{}0$\r\nEither $ f(2x)\\equal{}2f(x)$\r\nEither $ f(2x)\\equal{}\\minus{}f(x)\\equal{}f(\\minus{}x)$ and so $ 2x\\equal{}\\minus{}x$ since $ f(x)$ is injective and so $ x\\equal{}0$ and $ f(2x)\\equal{}2f(x)$\r\nQ.E.D.\r\n\r\n5.2) $ f(x^3)\\equal{}f(x)^3$ and, as a consequence, $ f(1)\\equal{}1$ or $ f(1)\\equal{}\\minus{}1$\r\n$ P(x,x)$ $ \\implies$ $ f(2x^3)\\equal{}2f(x)^3$. But $ f(2x^3)\\equal{}2f(x^3)$\r\nQ.E.D.\r\n\r\n5.3) $ f(nx)\\equal{}nf(x)$ $ \\forall x$, $ \\forall n\\in \\mathbb N$\r\nThis is true for $ n\\equal{}1$\r\nSuppose it's true for $ n$. Then $ P(x,nx)$ $ \\implies$ $ f(n(n\\plus{}1)x^3)\\equal{}f(nx)f(x)(f(nx)\\plus{}f(x))$\r\n$ \\implies$ $ nf((n\\plus{}1)x^3)\\equal{}n(n\\plus{}1)f(x)^3$\r\n$ \\implies$ $ f((n\\plus{}1)x^3)\\equal{}(n\\plus{}1)f(x^3)$\r\n$ \\implies$ $ f((n\\plus{}1)x)\\equal{}(n\\plus{}1)f(x)$ $ \\forall x$\r\nQ.E.D\r\n\r\n5.4) $ f(x)\\equal{}x$ $ \\forall x\\in\\mathbb Q$ or $ f(x)\\equal{}\\minus{}x$ $ \\forall x\\in\\mathbb Q$\r\nWe have $ f(nx)\\equal{}nf(x)$ $ \\forall n\\in\\mathbb N$\r\nSince we also have $ f(\\minus{}x)\\equal{}\\minus{}f(x)$, we get $ f(nx)\\equal{}nf(x)$ $ \\forall n\\in\\mathbb Z$\r\n$ f(p\\frac xq)\\equal{}pf(\\frac xq)$ and $ f(x)\\equal{}f(q\\frac xq)\\equal{}qf(\\frac xq)$ $ \\implies$ $ f(ax)\\equal{}af(x)$ $ \\forall a\\in\\mathbb Q$\r\nAnd since $ f(1)\\equal{}1$ or $ f(1)\\equal{}\\minus{}1$, we get the result.\r\n\r\n6) Non constant solutions are such that $ f(x)\\equal{}x$ $ \\forall x\\in\\mathbb R$ or $ f(x)\\equal{}\\minus{}x$ $ \\forall x\\in\\mathbb R$\r\nIf $ f(x)$ is solution, $ \\minus{}f(x)$ is solution too. So Wlog consider solutions where $ f(1)\\equal{}1$ and $ f(x)\\equal{}x$ $ \\forall x\\in\\mathbb Q$\r\n\r\nLet $ a,b$ such that $ b^4\\plus{}4ab\\geq 0$. Then the equation $ xb(x\\plus{}b)\\equal{}a$ has solutions and, for $ x$ such that $ xb(x\\plus{}b)\\equal{}a$, we have :\r\n$ f(a)\\equal{}f(x)f(b)(f(x)\\plus{}f(b))$ and so the equation $ X^2f(b)\\plus{}Xf(b)^2\\minus{}f(a)$ has solutions and so $ f(b)^4\\plus{}4f(a)f(b)\\geq 0$\r\n\r\nSo $ b^4\\plus{}4ab\\geq 0$ $ \\implies$ $ f(b)^4\\plus{}4f(a)f(b)\\geq 0$\r\nFrom this, we can get three important conclusions :\r\n6.1) $ b>0$ $ \\implies$ $ f(b)>0$\r\nSuppose $ b>0$ such that $ f(b)<0$. It's then easy to find a rational $ a>0$ great enough to have $ b^4\\plus{}4ab\\geq 0$ and $ f(b)^4\\plus{}4f(a)f(b)\\equal{}f(b)^4\\plus{}4af(b)< 0$\r\n\r\n6.2) $ f(x)\\leq x$ $ \\forall x>0$\r\nConsider now $ a>0$ and $ b<0\\in\\mathbb Q$ such that $ \\minus{}\\frac{b^3}{4}\\geq a$. Then $ b^4\\plus{}4ab\\geq 0$ and so $ f(b)^4\\plus{}4f(a)f(b)\\geq 0$ and so $ \\minus{}\\frac{b^3}{4}\\geq f(a)$ (since $ f(b)\\equal{}b$).\r\n\r\nSo $ \\minus{}\\frac{b^3}{4}\\geq a$ $ \\implies$ $ \\minus{}\\frac{b^3}{4}\\geq f(a)$ and, since we can always find rational $ b$ such that $ \\minus{}\\frac{b^3}{4}$ is as near of $ a$ as we want : $ f(a)\\leq a$ $ \\forall a>0$\r\n\r\n6.3) $ f(x)\\geq x$ $ \\forall x>0$\r\nConsider now $ b>0$ and $ a<0\\in\\mathbb Q$ such that $ \\minus{}\\frac{b^3}{4}\\leq a$. Then $ b^4\\plus{}4ab\\geq 0$ and so $ f(b)^4\\plus{}4f(a)f(b)\\geq 0$ and so $ f(b)^4\\plus{}4af(b)\\geq 0$. So, since $ f(b)>0$ (see 6.1) : $ \\minus{}\\frac{f(b)^3}{4}\\leq a$\r\n\r\nAnd so, since we can always find rational $ a$ such that $ a$ is as near of $ \\minus{}\\frac{b^3}{4}$ as we want : $ \\minus{}\\frac{f(b)^3}{4}\\leq \\minus{}\\frac{b^3}{4}$ and so $ f(b)^3\\geq b^3$ and so $ f(b)\\geq b$\r\n\r\n\r\nSo 6.2 and 6.3 imply $ f(x)\\equal{}x$ $ \\forall x>0$ and so $ f(x)\\equal{}x$ $ \\forall x$\r\n\r\n7) Synthesis of solutions :\r\n$ f(x)\\equal{}0$\r\n$ f(x)\\equal{}\\frac{1}{\\sqrt 2}$\r\n$ f(x)\\equal{}\\minus{}\\frac{1}{\\sqrt 2}$\r\n$ f(x)\\equal{}x$\r\n$ f(x)\\equal{}\\minus{}x$" +} +{ + "Tag": [ + "calculus", + "trigonometry", + "function", + "inequalities", + "limit", + "real analysis", + "real analysis unsolved" + ], + "Problem": "\"If you ever meet a boastful mathematician, show him this limit.. it will keep him quiet for a while..\" -- my calculus lecturer..\r\n\r\nI've been given this to solve as a challenge:\r\n\r\nlim (as x --> 0) of\r\n\r\ntan(x).sin(x) - sin(x).tan(x)\r\n--------------------------------------\r\ntan-1(x).sin-1(x) - sin-1(x).tan-1(x)\r\n\r\n[Sorry for the crappy layout.\r\nIt's the limit as x tends to zero of (the top line divided by the bottom line).\r\nThe bottom line are tan and sine inverses..]\r\n\r\nI can't do it but I'm guessing to use L'Hopital's Rule. I had a go at it, but it got far too complicated and I didn't see anything which would deviate the expression from cancelling to 0/0.\r\nMathematica claims it's an indeterminate limit, but this only goes to show that the human mind is far more sophisticated than Mathematica. It IS determinate.. so any ideas?", + "Solution_1": ":rotfl: Ok, of course it will keep it quite a while, because he will surely not try to find a solution to this. Actually, who would? But the approach is straightforward: just develop each function in Taylor series and take sufficiently many terms from each development so that you can avoid cancelling all terms. Basically, both the numerator and denominator will be of the form $ax^{n}+o(x^{n})$ for some $n$ and some nonzero $a$. Then you know what you have to do. I know that the computations can be extremely messy, but that is no longer a matter of inteligence or mathematics, it is just a matter of brute force (see Muierhead and the way some people kill inequalities, for instance...).", + "Solution_2": "We haven't done Taylor series yet so I'm sure I'd make a mess of it.. it's the generalised form or the Maclaurin series?\r\n\r\nSo there is no other way of solving this? Would you be able to give me a head start on it then, using Taylor series? I could clean up the mess myself.", + "Solution_3": "[quote=\"phalaris\"]\ntan(x).sin(x) - sin(x).tan(x)\n--------------------------------------\ntan-1(x).sin-1(x) - sin-1(x).tan-1(x)\n[/quote]\r\nAre you sure that this is what you mean? You are basically asking $\\lim_{x \\to 0}\\frac00$ :huh: Maybe you got it wrong, or something like that...", + "Solution_4": "Yes, I'm 100% sure. I wouldn't have emphasized that something which looks so simple is actually a really tricky and challenging limit if I wasn't sure.\r\n\r\nI mean the idea is that 0/0 is an ambiguous notion when it comes to limits. (Likewise for inf/inf, 1^inf, etc..)\r\nOne may or may not tend to 0 faster than another or perhaps a different relationship might be established.\r\n\r\nThe lim (as x --> 0) sin(x)/x = 1 , even though sin(0) is zero. This can easily be proved using L'Hopital's Rule:\r\n\r\nd(sin(x))/dx = cos(x);\r\nd(x)/dx = 1;\r\n\r\n=> lim (as x --> 0) sin(x)/x = lim (as x --> 0) cos(x)/1 = 1\r\n\r\nC'mon people.. I need a solution! :D", + "Solution_5": "[quote]I mean the idea is that 0/0 is an ambiguous notion when it comes to limits. (Likewise for inf/inf, 1^inf, etc..)[/quote]\nIt's an abuse of notation! And like most of these, it can cause confusion. $\\frac00$ is not defined, but in calculus classes, one calls $\\lim_{x \\to a}\\frac{f(x)}{g(x)}$ a $\\frac00$ limit if $\\lim_{x \\to a}f(x) = \\lim_{x \\to a}g(x) = 0$ and $g(x) \\neq 0$ on a small enough neighbourhood of $a$.\n\n[quote=\"phalaris\"]C'mon people.. I need a solution! :D[/quote]\r\nTo what? Your limit is not correctly defined. Most people agree that $\\frac00$ is not defined (there are some who don't and try to give it a new \"value\" - there was some topic around here, I can't remember where); to work with the limit $\\lim_{x \\to 0}f(x)$, the function $f$ should be defined on $\\mathcal V(0) \\setminus \\left\\{ 0 \\right\\}$, where $\\mathcal V(0)$ is a neighbourhood of $0$.", + "Solution_6": "Ok, but then by the same argument the lim (as x --> 0) of sin(x)/x is not defined either..\r\n\r\nAnyway I don't want to have an argument, I just want someone to help me with a solution. If I don't get one here then I'll be happy to PM it to you if/when my lecturer reveals what it is.", + "Solution_7": "There's a pretty good chance that he means [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=101678]this problem[/url].\r\n\r\nphalaris: if this is what you meant, then you stated it incorrectly. Can you see that? There's a big difference between $\\sin x\\cdot\\tan x$ and $\\sin(\\tan x).$ The \"$\\frac00$\" of limits does not men \"always exactly zero\" over \"always exactly zero\", which is what you wrote - that isn't delicate or tricky, it's merely meaningless. But the problem I referenced above is delicate and tricky - and discussed before.", + "Solution_8": "Looking at what I have written down here I would have to say no.\r\n\r\nI doubt my maths lecturer is such a retard as to give us the wrong limit.\r\n\r\nThanks anyway. I'll ask him on tuesday and if he meant what he said, and if he gives me a solution then I will post it here.", + "Solution_9": "[quote=\"phalaris\"]I doubt my maths lecturer is such a retard as to give us the wrong limit.[/quote]\r\nIf I had to guess, I'd guess that he gave you the problem referenced, namely\r\n\\[\\lim_{x\\to0}\\frac{\\tan(\\sin x)-\\sin(\\tan x)}{\\tan^{-1}(\\sin^{-1}x)-\\sin^{-1}(\\tan^{-1}x)},\\]\r\nand that you're the one who copied it down incorrectly.", + "Solution_10": "[quote=\"Kent Merryfield\"][quote=\"phalaris\"]I doubt my maths lecturer is such a retard as to give us the wrong limit.[/quote]\nIf I had to guess, I'd guess that he gave you the problem referenced, namely\n\\[\\lim_{x\\to0}\\frac{\\tan(\\sin x)-\\sin(\\tan x)}{\\tan^{-1}(\\sin^{-1}x)-\\sin^{-1}(\\tan^{-1}x)}, \\]\nand that you're the one who copied it down incorrectly.[/quote]\r\n\r\nidea:when x->0 then\r\ntan(sin(x))-sin(tanx))=((x^7)/30)+o(x^7)", + "Solution_11": "That's true- but how did you calculate it? It's also in the linked thread, with a full explanation. My solution there has a typo, but it shouldn't be too hard to see past.", + "Solution_12": "[quote=\"phalaris\"]Looking at what I have written down here I would have to say no.\n\nI doubt my maths lecturer is such a retard as to give us the wrong limit.\n\nThanks anyway. I'll ask him on tuesday and if he meant what he said, and if he gives me a solution then I will post it here.[/quote]\r\n\r\nThen that would make me the retard, and rightly so.. :blush: \r\n\r\nThanks for all your input. I'll let you guys know on tuesday.", + "Solution_13": "[quote=\"Extremal\"][quote=\"Kent Merryfield\"][quote=\"phalaris\"]I doubt my maths lecturer is such a retard as to give us the wrong limit.[/quote]\nIf I had to guess, I'd guess that he gave you the problem referenced, namely\n\\[\\lim_{x\\to0}\\frac{\\tan(\\sin x)-\\sin(\\tan x)}{\\tan^{-1}(\\sin^{-1}x)-\\sin^{-1}(\\tan^{-1}x)}, \\]\nand that you're the one who copied it down incorrectly.[/quote]\n\nidea:when x->0 then\ntan(sin(x))-sin(tanx))=((x^7)/30)+o(x^7)[/quote]\r\nhttp://ivanisvili.narod.ru/Extr3.doc I mean taht it is not hard. the hard is to find taylor for tan(x). only this." +} +{ + "Tag": [ + "algebra", + "polynomial" + ], + "Problem": "what does roots of a polynomial mean in real life?", + "Solution_1": "This is a very vague question. You'd do well to ask what a polynomial means in \"real life,\" and the answer is that it's completely up to you. The tools of polynomial algebra are just that - mathematical [b]tools[/b]. They can be applied to real-world problems in a variety of ways. For example, mathematicians used to think that negative numbers - and later, imaginary numbers - had no meaning \"in real life,\" but of course negative numbers are a convenient way to keep track of, say, trade, and imaginary numbers have a variety of physical applications even if they don't \"mean\" anything \"in real life.\"", + "Solution_2": "i disagree with ur answer. i have a very good reason,proof.\r\nit took me 3 years to find out,but hard to explain in words.", + "Solution_3": "[quote=\"binomial_4eva\"]i have a very good reason,proof.[/quote]\r\n\r\nHow do you propose to [b]prove[/b] something about [b]meaning[/b] in the [b]real world[/b]?", + "Solution_4": "it is hard to explain it in here,too long to write, and plus my english is crap, if i prove, you might not undrestand me, you might even laugh at my proof.", + "Solution_5": "Too late, I'm already laughing." +} +{ + "Tag": [ + "inequalities", + "integration", + "calculus", + "real analysis", + "real analysis unsolved" + ], + "Problem": "$ (X,m,\\mu)$ is measure space with $ \\mu(X)\\equal{}1$. Suppose that $ 1\\leq r\\leq s\\leq \\infty$. \r\n\r\nShow that $ L^s(\\mu)\\subseteq L^r(\\mu)$ \r\nand the inclusion mapping $ j: L^s(\\mu)\\hookrightarrow L^r(\\mu): f\\rightarrow f$ is bounded.", + "Solution_1": "let $ f \\in L^r (\\mu )$\r\n\r\nthen $ X\\equal{}\\{x \\in X : |f(x)| \\le 1 \\} \\cup \\{x \\in X : |f(x)| > 1 \\}$\r\n\r\nthen \r\n\r\n$ \\int_X |f(x)|^s dx \\equal{} \\int_{\\{x \\in X : |f(x)| \\le 1 \\}} |f(x)|^s dx\\plus{} \\int_{\\{x \\in X : |f(x)| > 1 \\}} |f(x)|^s dx \\le$\r\n$ \\int_X |f(x)|^s dx\\plus{} \\int_{\\{x \\in X : |f(x)| > 1 \\}} |f(x)|^r dx \\le \\mu (X) \\plus{} \\int_X |f(x)|^r dx < \\infty$", + "Solution_2": "There is something fishy there. I think it is just a typo, though. Should the first integral after the first inequality sign be $ \\int_X 1^s dx$?", + "Solution_3": "yes. srry typo" +} +{ + "Tag": [ + "geometry" + ], + "Problem": "An equiangular octagon has four sides of length $ 1$ and four sides of length $ \\frac{\\sqrt{2}}{2}$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?\r\n\r\n$ \\textbf{(A)}\\ \\frac{7}{2}\\qquad \r\n\\textbf{(B)}\\ \\frac{7\\sqrt{2}}{2}\\qquad \r\n\\textbf{(C)}\\ \\frac{5 \\plus{} 4\\sqrt{2}}{2}\\qquad \r\n\\textbf{(D)}\\ \\frac{4 \\plus{} 5\\sqrt{2}}{2}\\qquad \r\n\\textbf{(E)}\\ 7$", + "Solution_1": "This octagon could be thought of as a square with $ \\frac{1}{2}$ inch cut off at each corner. the square would have side length $ 2$. Each of the triangular portions cut off have area $ \\frac{1}{2}\\times\\frac{1}{4}\\equal{}\\frac{1}{8}$. $ 4\\times\\frac{1}{8}\\equal{}\\frac{1}{2}$. The answer is the area of the square minus the total area cut off, which is $ 4\\minus{}\\frac{1}{2}\\equal{}\\frac{7}{2}$. The answer is $ \\fbox{A}$.", + "Solution_2": "[hide=\"Another Solution\"]\nWe can partition the octagon into five squares and four half squares, each with side length $ \\frac{\\sqrt{2}}{2}$, so its area is $ \\left (5 \\plus{} 4 \\cdot \\frac{1}{2} \\right ) \\left (\\frac{\\sqrt{2}}{2} \\right )^2\\equal{}\\frac{7}{2}$, or answer choice $ \\boxed{\\textbf{A}}$.\n[/hide]" +} +{ + "Tag": [], + "Problem": "Prove that for n a positive integer , 17 + 19.(8^n) is never a prime number.", + "Solution_1": "[quote=\"Michael Niland\"]Prove that for n a positive integer , 17 + 19.(8^n) is never a prime number.[/quote]\r\n\r\n* n even:\r\nModulo 3:\r\n17 + 19*(8^n) = 2 + 1*(-1^n) = 3 = 0 so 17 + 19*(8^n) is divisible by 3.\r\n\r\n*n odd?", + "Solution_2": "For $n\\equiv 1 \\mod 4$ looking modulo $13$ and for $n\\equiv 3 \\mod 4$ looking modulo $5$ works well." +} +{ + "Tag": [ + "vector", + "geometry proposed", + "geometry" + ], + "Problem": "Let A(1)A(2)..A(n) be a convex poligon, and M a point in The interior of the poligon. Find the minimum m natural, for which the convex poligon B(1)B(2)...B(m+n) has \"G - weight point\" in M.", + "Solution_1": "In fact, {B(1),B(2),...B(n+m)} includes {A(1),A(2)...,A(n)}\r\nand B(i)<>B(j) for any i<>j.\r\nSorry !! :)", + "Solution_2": "What is G?\r\nAs I remember, It is a point in the plane such that the sum of all vector A(i)G \r\nis 0. For that point in the plane, calculate the sum of all the vectors A(i)M and ull get another vector. Then, it shuold be easier." +} +{ + "Tag": [ + "algebra unsolved", + "algebra" + ], + "Problem": "caculate : \r\n $ \\sqrt{111111...111\\minus{}2222..222}$\r\n where : have 1998 number 1 and 999 number 2", + "Solution_1": "It is \r\n$ \\frac{10^n\\minus{}1}{3}$", + "Solution_2": "Let $ N\\equal{}\\underbrace{111\\cdots 1}_{999}$. Then we have\r\n\\[ \\sqrt{N(10^{999}\\plus{}1)\\minus{}2N}\\equal{}\\sqrt{N(10^{999}\\minus{}1)}\\equal{}\\sqrt{N(9N)}\\equal{}3N\\]" +} +{ + "Tag": [ + "search", + "floor function", + "combinatorics proposed", + "combinatorics" + ], + "Problem": "A non-empty set $ S$ of positive integers is said to be [i]good[/i] if there is a coloring with $ 2008$ colors of all positive integers so that no number in $ S$ is the sum of two different positive integers (not necessarily in $ S$) of the same color. Find the largest value $ t$ can take so that the set $ S\\equal{}\\{a\\plus{}1,a\\plus{}2,a\\plus{}3,\\ldots,a\\plus{}t\\}$ is good, for any positive integer $ a$.\r\n\r\n[hide=\"P.S.\"]I have the feeling that I've seen this problem before, so if I'm right, maybe someone can post some links...[/hide]", + "Solution_1": "http://www.mathlinks.ro/viewtopic.php?search_id=954790132&t=149160", + "Solution_2": "This really isn't some very hard problem, I'm surprised that nobody has posted some solution already...\r\nAnswer is $ t \\equal{} 2\\cdot 2007 \\equal{} 4014$.\r\nLook at the set $ M \\equal{} \\{\\frac {a}2, \\frac {a }2 \\plus{} 1, ..., \\frac {a}2 \\plus{} 2008\\}$ for some even $ a$. This set contains $ 2009$ elements, so two of them have the same color. In other hand, if $ b,c\\in M$ with $ b \\not \\equal{} c$ then $ a \\plus{} 1 \\leq b \\plus{} c \\leq a \\plus{} 1 \\plus{} 2\\cdot 2007$, so $ t$ must be less then $ 1 \\plus{} 2\\cdot 2007$ because there is no good set $ S\\equal{}\\{a\\plus{}1, a\\plus{}2, ..., a\\plus{}2\\cdot 2007 \\plus{} 1\\}$ when $ a$ is even.. \r\nWhen $ t \\equal{} 2\\cdot 2007$ we color the numbers like this: numbers $ 1,2,...,\\lfloor \\frac {a \\plus{} 1}2 \\rfloor$ have color $ 1$, number $ \\lfloor \\frac {a \\plus{} 1}2 \\rfloor \\plus{} 1$ is in color $ 2$, $ \\lfloor \\frac {a \\plus{} 1}2 \\rfloor \\plus{} 2$ have color $ 3$, ..., $ \\lfloor \\frac {a \\plus{} 1}2 \\rfloor \\plus{} 2006$ have color $ 2007$, and numbers from $ \\lfloor \\frac {a \\plus{} 1}2 \\rfloor \\plus{} 2007$ and grater have color $ 2008$. You see that if you pick two different numbers with same color, their sum is either smaller then $ a \\plus{} 1$ or bigger then $ a \\plus{} 2\\cdot 2007$, and we're done :)\r\n\r\nBye" +} +{ + "Tag": [ + "geometry", + "AMC", + "AIME" + ], + "Problem": "Does anyone have any specific advice on how to improve your geometry, especially at the aime level problems 6-12? (i.e more than, do geometry problems.) Thanks!", + "Solution_1": "Know theorms(stewarts, ptolomy,...), and be good with similar triangles", + "Solution_2": "well intro to geometry is awesome but much easier than what you're looking for\r\ni think that intermediate geometry is coming out in a few months", + "Solution_3": "For this level, knowing theorems beyond those taught in the normal school curriculum will have very limited value. Instead, try to look for efficient ways to calculate various lengths/angles/areas. The AIME is all about calculation, and most problems will fall after doing enough of this in an intelligent enough way. Also, the great thing about geometry is that you'll be sure to improve with practice." +} +{ + "Tag": [ + "analytic geometry", + "graphing lines", + "slope", + "combinatorics proposed", + "combinatorics" + ], + "Problem": "In the future city Baltic Way there are sixteen hospitals. Every night exactly four of them must be on duty for emergencies. Is it possible to arrange the schedule in such a way that after twenty nights every pair of hospitals have been on common duty exactly once?", + "Solution_1": "[hide]No. Each night, six pairs of hospitals share common duty. Between 16 hospitals, there are 120 (16*15/2) necessary pairs. Fot twenty nights, there would be 120 pairs if there were no overlaps where a pair was used twice. However, a pair must be used on two nights. Assume the opposite. If no overlap was to be made, this could be kept up for five nights, and each of the twenty would belong to one pair of four. After this, the sixth group of four would have to include four members from other groups.[/hide]", + "Solution_2": "[quote=\"Icosahedroid\"][hide]No. Each night, six pairs of hospitals share common duty. Between 16 hospitals, there are 120 (16*15/2) necessary pairs. Fot twenty nights, there would be 120 pairs if there were no overlaps where a pair was used twice. However, a pair must be used on two nights. Assume the opposite. If no overlap was to be made, this could be kept up for five nights, and each of the twenty would belong to one pair of four. After this, the sixth group of four would have to include four members from other groups.[/hide][/quote]\r\nYour proof is flawed. Your conclusion is \"After this, the sixth group of four would have to include four members from other groups.\", but this is not a contradiction to your assumption!", + "Solution_3": "If 1 means a hospital must be on duty for emergencies and zero means conversely that a hospital must not be on duty for emergencies then two solutions are\r\n\r\n1111000000000000\r\n0000111100000000\r\n0000000011110000\r\n0000000000001111\r\n1000100010001000\r\n0100010001000100\r\n0010001000100010\r\n0001000100010001\r\n0001001001001000\r\n0100000100101000\r\n0010010000011000\r\n0010000110000100\r\n0001100000100100\r\n1000001000010100\r\n0001010010000010\r\n1000000101000010\r\n0100100000010010\r\n0100001010000001\r\n0010100001000001\r\n1000010000100001\r\n\r\nor for example\r\n\r\n1111000000000000\r\n0000111100000000\r\n0000000011110000\r\n0000000000001111\r\n1000100010001000\r\n0100010001000100\r\n0010001000100010\r\n0001000100010001\r\n0010000101001000\r\n0001010000101000\r\n0100001000011000\r\n0001001010000100\r\n1000000100100100\r\n0010100000010100\r\n0100000110000010\r\n0001100001000010\r\n1000010000010010\r\n0010010010000001\r\n1000001001000001\r\n0100100000100001", + "Solution_4": "That's more like it! :)\r\nI found out through my father that this situation actually corresponds to the one of a speedway championsship. (With 16 drivers, 20 heats with 4 in each, and everyone meets the other exactly once)\r\n\r\nDid you find a pattern, or was it just pure \"guesswork\"?", + "Solution_5": "I found the solutions from a Finnish math discussion forum. It was said that the original solver wrote some Fortran code.", + "Solution_6": "Here's a different way of finding a solution if you know a bit of algebra:\r\n\r\nLet $ Q$ be the finite field of $ 4$ elements. We can associate the $ 16$ hospitals with the $ 16$ points $ (x,y)$, where $ x$ and $ y$ are elements of $ Q$. \r\n\r\nThis $ 4 \\times 4$ grid contains $ 20$ lines corresponding to equations $ ax \\plus{} by \\equal{} c$. (There are $ 5$ possible \"slopes\": vertical, and one for each element of the field, and each slope has 4 parallel lines). Each pair of points is on exactly one line.", + "Solution_7": "[quote=\"kevinatcausa\"]Here's a different way of finding a solution if you know a bit of algebra:\n\nLet $ Q$ be the finite field of $ 4$ elements. We can associate the $ 16$ hospitals with the $ 16$ points $ (x,y)$, where $ x$ and $ y$ are elements of $ Q$. \n\nThis $ 4 \\times 4$ grid contains $ 20$ lines corresponding to equations $ ax \\plus{} by \\equal{} c$. (There are $ 5$ possible \"slopes\": vertical, and one for each element of the field, and each slope has 4 parallel lines). Each pair of points is on exactly one line.[/quote]\r\nWow! Thanks a lot :)\r\nI don't know why you want to let $ Q$ be the \"finite field of $ 4$ elements\" instead of just $ \\{1,2,3,4\\}$ though. Seems less intuitive to me :P", + "Solution_8": "If you just tried to work on the grid modulo $ 4$, the \"line\" going through $ (0,0)$ and $ (1,2)$ (which you would naturally want to extend to $ (2, 2 \\plus{} 2) \\equal{} (2, 0)$ and $ (3,2)$ ) and the line through $ (0,0)$ and $ (1,0)$ intersect in more than one point.\r\n\r\nEssentially this is because you can't do division modulo $ 4$. The line through $ (0,0)$ and $ (2,0)$ should have a slope satisfying $ 2m \\equal{} 0$, but this doesn't uniquely determine $ m$. By moving to a field, we guarantee that this equation has a unique solution for any two points not sharing the same $ x$ coordinate.\r\n\r\nOn a slight tangent: For any $ q$ which is a prime power, the same argument gives an arrangement with $ q^2$ hospitals (points) and $ q^2 \\plus{} q$ shifts (lines). Such a structure is known as an affine plane. The argument breaks down when $ q$ is not a prime power, because then there are no fields of order $ q$. It's a rather infamous open problem to show that if $ q$ is not a prime power, then you can [b]not[/b] find such an arrangement with $ q^2$ points and $ q^2 \\plus{} q$ lines. The most recently solved case was $ q \\equal{} 10$, which required an extensive computer search. Even $ q \\equal{} 12$ remains open.", + "Solution_9": "This is an easy application of balanced block designs.It is well-known that a projective plane of order $4$ exists and that a affine plane if order $4$ is the reminder design of a projective plane of order $4$ so we have a $2-(16,4,1)$ design which is exactly what the problem is asked us to do." +} +{ + "Tag": [ + "induction", + "function", + "Putnam" + ], + "Problem": "This problem is old, but very good for this level,\r\n\r\nProve that the following expression holds for all positive integers $n\\in\\mathbb{Z}^+$,\r\n\\[\\sum_{k=0}^{n}{n\\choose k}^2={2n\\choose n}\\]\r\n\r\nThere is a very simple combinatorical argument, but there is also an induction kind of algebra solution that works fine.\r\n\r\nBest,", + "Solution_1": "[hide=\"proof\"]this is a simple method deriving from what's called block walking (from AoPS book v.2)\n\n$\\displaystyle\\sum_{k=0}^{\\infty}\\displaystyle{n\\choose k}^2 = \\displaystyle{2n\\choose n}$.\n\n\nWalking from $\\displaystyle{0\\choose 0}$ to $\\displaystyle{2n\\choose n}$, we must pass through the nth row, namely $\\displaystyle{n\\choose k}$. \n\nfrom $\\displaystyle{n\\choose k}$, to go to $\\displaystyle{2n\\choose n}$, we have ${\\displaystyle{n\\choose \\left(n - k \\right)}}$, because out of n ways remaining to go, there are $n - k$ ways to go to the right branches. thus, number of ways passing through $\\displaystyle{n\\choose k}$ is \n\n$\\displaystyle{n\\choose k}\\displaystyle{n\\choose \\left(n - k \\right)}$ \n$ = \\displaystyle{n\\choose k}\\displaystyle{n\\choose k}$\n$ = \\displaystyle{n\\choose k}^2$\n\nbecause there can be any value from 0 to n for k, we can say that:\n\n$\\displaystyle{2n\\choose n} = \\displaystyle{n\\choose 0}^2 + \\displaystyle{n\\choose 1}^2 + \\displaystyle{n\\choose 2}^2 + \\displaystyle{n\\choose 3}^2 + ... + \\displaystyle{n\\choose n}^2$, \n\nthus finishing the proof. [/hide]\r\n\r\nI have read this too many times, i memorize the whole steps now.\r\n\r\nin fact, i have a similar problem you can prove:\r\n\r\nProve $\\displaystyle\\sum_{k=0}^m\\displaystyle{n\\choose k}{n-k\\choose m-k} = 2^m {n\\choose m}$.\r\n\r\nSOURCE, AoPS book2", + "Solution_2": "Generating functions:\r\n\r\n$(1+x)^{2n}$ is a generating function for the $LHS$, $(1+x)^n(1+x)^n$ is a generating function for the $RHS$ (if you take the Cauchy product).\r\n\r\nAs for the other argument which I prefer, let $X$ be a set with $2n$ objects, and say that $n$ of the objects are black and the other $n$ are white. Then the way of choosing $2n$ objects is the number of ways of choosing $0$ black, $n$ white, $1$ black, $n-1$ white, etc, etc. Use the fact that choosing $k$ objects from $n$ is equal to the number of ways of choosing $n-k$ from $n$, and we're done.", + "Solution_3": "nice proof. i didnt even see that. that is a nice way of applying to solve.", + "Solution_4": "Does anyone knows where you can find a good material about generating functions? I am interested on it, I have to prepare some classes for the fall, for the Putnam preparation!\r\n\r\nBest," +} +{ + "Tag": [ + "function", + "limit", + "algebra proposed", + "algebra" + ], + "Problem": "Suppose that $f: [0.\\infty)\\rightarrow [0,\\infty)$ be a continuous function that not equal to zero and for every $x\\in [0,\\infty) , f(f(x))=(x^2+x+1)f(x)$. Find the limitations\r\n\r\na)$\\lim_{x\\rightarrow 0} \\frac{f(x)}{x}$\r\n\r\nb)$\\lim_{x\\rightarrow \\infty} \\frac{f(x)}{x}.$", + "Solution_1": "First we can prove that $f(0)=0$. For this if we put $x=0$, we\u00b4ll obtain $f(f(0))=f(0)$. And so, we can afirm, by reduction to absurd, that $f(0)=k$ and $k>0$. Now we can considerate a function $g$ such that $g(x)=f(x)-x$. And so we\u00b4ll have $f(g(x)+x)=(x^2+x+1)(g(x)+x)$ $\\to$ $x+g(x)+g(x+g(x))=(x^2+x+1)(g(x)+x)$, and in this last equalitie we can put $x=k$, and so we\u00b4ll have $k=(k^2+k+1)k$, and hence $k=0$. That\u00b4s an absurd and so we have $f(0)=0$.\r\nNow considerate $h$, such that $h(x)=\\frac{f(x)}{x}$ and so we\u00b4ll have that $f(xh(x))=(x^2+x+1)xh(x)$ $\\to$ $xh(x)h(xh(x))=(x^2+x+1)xh(x)$. But $f$ is never zero for all $x>0$ and so $xh(x) > 0$. So we have $h(f(x))=(x^2+x+1)$. And now since both $f$ and $g$ are continuous functions we\u00b4ll have that $\\lim_{x\\to 0+}h(f(x))=\\lim_{x\\to 0+}(x^2+x+1)$ $\\to$ $\\lim_{x\\to 0+}h(x)=1$. We can obtain too that $\\lim_{x\\to +\\infty}h(x)=+\\infty$" +} +{ + "Tag": [ + "geometry" + ], + "Problem": "Do radii of intersecting circles form right angles at the points of intersection?\r\nAlso, to put it another way, must radii of intersecting circles be tangents to the opposing circles at the points of intersection?", + "Solution_1": "Not necessarily. Draw four radii in two intersecting circles from the centers to the intersection points. Then you have a rhombus. Note that the angles at the centers of the circles have measures equal to the measure of the arcs between the intersection points, and the other angles are just supplements (if the circles are congruent: in any case, the other angles are congruent, and everything adds to $2\\pi$ or 360 degrees). Now...there are an infinite number of possibilities for the measure of these arcs, correct?", + "Solution_2": "One definite counterexample is if you just drew two circles that intersect at one point.", + "Solution_3": "[quote=\"mathpr\"]Do radii of intersecting circles form right angles at the points of intersection?\nAlso, to put it another way, must radii of intersecting circles be tangents to the opposing circles at the points of intersection?[/quote]\r\n\r\nWhen the above condition is indeed satisfied, we have what we call orthogonal circles... :)" +} +{ + "Tag": [ + "algebra", + "polynomial" + ], + "Problem": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b2\u03ac\u03b6\u03c9 \u03bc\u03b9\u03b1 \u03b1\u03c3\u03ba\u03b7\u03c3\u03bf\u03cd\u03bb\u03b1, \u03b1\u03bb\u03bb\u03ac \u03af\u03c3\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae \u03ba\u03b1\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03be\u03b1\u03bd\u03b1\u03bc\u03c0\u03b5\u03af \u03c3\u03c4\u03bf forum, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03c3\u03cc\u03c1\u03c1\u03c5 \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03cc :oops: \r\n\r\nN\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03cd\u03bd \u03bf\u03b9 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03b5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b7\u03c2 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7\u03c2\r\n\r\n$ \\frac {x^7 \\minus{} 1}{x \\minus{} 1} \\equal{} y^5 \\minus{} 1$", + "Solution_1": "[url]http://www.mathlinks.ro/viewtopic.php?p=780855#780855[/url]", + "Solution_2": "\u03a4\u03b7\u03bd \u03b5\u03af\u03c7\u03b5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf \u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u03bf \u03a0\u03b1\u03c0\u03c0\u03ad\u03bb\u03b7\u03c2 \u03c0\u03b1\u03bb\u03b9\u03cc\u03c4\u03b5\u03c1\u03b1 . \u039d\u03b1 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03c9 mim \u03cc\u03c4\u03b9 \u03c4\u03b7 \u03bb\u03cd\u03bd\u03b5\u03b9\u03c2 \u03bc\u03b5 cyclotomic ? :P", + "Solution_3": "Sil \u03ba\u03b1\u03b9 \u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c4\u03b5 \u03c4\u03b1 cyclotomic polynomials \u03b5 ? \u039c\u03bd \u03be\u03b5\u03c7\u03ac\u03c3\u03b5\u03c4\u03b5 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03b5\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7.....Thanks.....", + "Solution_4": "[quote=\"silouan\"]\u03a4\u03b7\u03bd \u03b5\u03af\u03c7\u03b5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf \u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u03bf \u03a0\u03b1\u03c0\u03c0\u03ad\u03bb\u03b7\u03c2 \u03c0\u03b1\u03bb\u03b9\u03cc\u03c4\u03b5\u03c1\u03b1 . \u039d\u03b1 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03c9 mim \u03cc\u03c4\u03b9 \u03c4\u03b7 \u03bb\u03cd\u03bd\u03b5\u03b9\u03c2 \u03bc\u03b5 cyclotomic ? :P[/quote]\r\n\r\nNice guess :wink: \u0392\u03ad\u03b2\u03b1\u03b9\u03b1, \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c4\u03b7\u03bd \u03ad\u03b2\u03b1\u03bb\u03b1 :P \r\n\u039f\u03bb\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c7\u03c1\u03ae\u03c3\u03b9\u03bc\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b9\u03c2 \u039f\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2, \u03ba\u03b1\u03b9 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03b1 :wink:\r\n\r\n\u03a3\u03cc\u03c1\u03c1\u03c5 \u03c0\u03bf\u03c5 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ad\u03c7\u03b5\u03b9 \u03be\u03b1\u03bd\u03b1\u03bc\u03c0\u03b5\u03af... (\u03b7 \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03ad\u03bd\u03b1 \u03c6\u03b1\u03b9\u03bd\u03cc\u03c4\u03b1\u03bd \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae)", + "Solution_5": "\u03ac\u03bb\u03bb\u03b5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=920822#p920822]\u03b5\u03b4\u03ce[/url] \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03c9" +} +{ + "Tag": [ + "function", + "algebra unsolved", + "algebra" + ], + "Problem": "let $ f$ and $ g$ be two real function with real variante st: $ f < g$ proof that there is a set $ M$ st: $ \\forall (x,y) \\in M^{2}$ we have : $ f(x) < g(y)$", + "Solution_1": "[quote=\"othman\"]let $ f$ and $ g$ be two real function with real variante st: $ f < g$ proof that there is a set $ M$ st: $ \\forall (x,y) \\in M^{2}$ we have : $ f(x) < g(y)$[/quote]\r\n\r\nMaybe I've not well understood the question, but $ M=\\{a\\}$ for any real $ a$ seems to fit.", + "Solution_2": "oh i'm sorry :blush: i forget to mention that the set $ M$ must be infinit , i think now the question ins't easy :D", + "Solution_3": "no one can solve it ? :rotfl: \r\n \r\n we have $ \\forall x, f(x)2$, we have $|\\cos \\phi|\\in\\{\\frac 12,1\\}$, as required.", + "Solution_2": "I am confused, doesn't algebraic integer just mean its a root of an equation? Or does it mean something else ?", + "Solution_3": "It means that it's a root of a monic polynomial with integral coefficients.", + "Solution_4": "ah thanks a lot! I understand now." +} +{ + "Tag": [ + "algebra proposed", + "algebra" + ], + "Problem": "$ a_1$ and $ a_2$ are positive integers less than $ 1000$. Define $ a_n \\equal{} min\\{|ai \\minus{} aj| : 0 < i < j < n\\}$. Show that $ a_{21} \\equal{} 0$.", + "Solution_1": "It is easy to see that $a_{2}\\ge 3a_4$ and $a_{n+2}\\le \\frac{a_n}{2}$ for all $n>2$.\nAs a result, $a_{21}=0$" +} +{ + "Tag": [ + "linear algebra", + "linear algebra unsolved" + ], + "Problem": "If $ AB\\equal{}BA\\equal{}0$ , show that for some integer $ k$, $ rank(A^k\\plus{}B^k)\\equal{}rank(A^k)\\plus{}rank(B^k)$.", + "Solution_1": "We can take $ k\\equal{}n$ where $ n$ is the $ A$-dimension.\r\n$ A^nB^n\\equal{}B^nA^n\\equal{}0$ and $ rank(A^{2n})\\equal{}rank(A^n)$.\r\nWe conclude with http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1630810#1630810" +} +{ + "Tag": [ + "calculus", + "derivative", + "inequalities unsolved", + "inequalities" + ], + "Problem": "2y^2 - x^2 = 1\r\n\r\nHow to get the min of\r\nf(x,y) = x-2y", + "Solution_1": "[quote=\"airrafer\"]2y^2 - x^2 = 1\n\nHow to get the min of\nf(x,y) = x-2y[/quote]\r\nLet $ x \\minus{} 2y \\equal{} m.$ Hence, $ x \\equal{} m \\plus{} 2y$ and the equation $ 2y^2 \\minus{} (m \\plus{} 2y)^2 \\equal{} 1$ has roots. :wink:", + "Solution_2": "[quote=\"airrafer\"]2y^2 - x^2 = 1\n\nHow to get the min of\nf(x,y) = x-2y[/quote]\r\n\r\n$ (x\\minus{}2y)^2 \\equal{} (2y^2 \\minus{} x^2) \\plus{} 2(x \\minus{} y)^2 \\ge 1$\r\n\r\nso we get $ x \\minus{} 2y \\ge 1$ or $ x \\minus{} 2y \\le \\minus{} 1$, on max and min... :wink:", + "Solution_3": "Can we use derivative?\r\nSet $ x \\equal{} \\sqrt {2y^2 \\minus{} 1}$ and get $ f(y) \\equal{} \\sqrt {2y^2 \\minus{} 1} \\minus{} 2y$ then $ f'(y) \\equal{} 0$ gives result." +} +{ + "Tag": [ + "calculus", + "limit", + "trigonometry", + "integration", + "function", + "calculus computations" + ], + "Problem": "(This is a short quiz. It's supposed to be easy. The use of L'H\u00f4pital is not explicitly forbidden - but if you make mistakes while using it, you get no sympathy.)\r\n\r\n1. Find $\\lim_{x\\to0}\\frac{\\cos x-e^{-x^{2}}}{x^{2}}$.\r\n\r\n2. Find the power series centered at zero for $\\int_{0}^{x}\\frac{dt}{1-t^{2}}$.\r\n\r\nWhat is the radius of convergence for this series?", + "Solution_1": "1. $\\frac{\\cos x-e^{-x^{2}}}{x^{2}}=\\frac{\\cos x-1}{x^{2}}+\\frac{1-e^{-x^{2}}}{x^{2}}$\r\n\r\n2. $\\int_{0}^{x}\\frac{dt}{1-t^{2}}=\\sum_{0}^{\\infty}\\frac{x^{2n+1}}{2n+1}$\r\nand the radius is 1.", + "Solution_2": "If you were in my class, I would expect you to actually state the answer on #1. :)", + "Solution_3": "Then let me go on for him/her. For $\\lim_{x \\to 0}$\r\n\r\n$\\frac{\\cos(x)-1}{x^{2}}=-\\cos(x)/2 =-1/2$\r\n\r\nand $\\frac{1-e^{-x^{2}}}{x^{2}}= \\frac{1-(1-x^{2}+x^{4}/2-x^{6}/3! ...)}{x^{2}}= 1-x^{2}/2+x^{4}/3!-... = 1$\r\n\r\nso it's $1/2$.\r\n\r\nHowever, it'd be far shorter if we make a single L'Hopital and plug 0.\r\n$\\frac{-\\sin(x)+2x e^{-x^{2}}}{2x}=-1/2+1$", + "Solution_4": "That's a double L'Hopital, though.", + "Solution_5": "Interesting how you get clusters of the same mistake for no good reason (and turn your suspicions off - I know where people were sitting and it's not that). The multiple mistakes of the day are $\\frac32$ on #1 and an alternating-sign power series (in particular, the arctan series) on #2.\r\n\r\nIf you want to get cute about it, that function in #2 is arctanh, the inverse hyperbolic tangent.\r\n\r\nIf I'd made the first problem $\\lim_{x\\to0}\\frac{\\cos x-e^{x^{2}/2}}{x^{4}},$ the success rate would have been significantly lower.", + "Solution_6": "for that new problem, couldn't one use taylor expansions up til the fourth degree so they would cancel with the denominator\r\nlim x-> 0 ((1-x^2/2 + x^4/24) - (1- x^2/2 + x^4/8))/x^4\r\nlim x-> 0 (-x^4/12)/(x^4) = -1/12", + "Solution_7": "Yes, that's the method and that's the answer. It shouldn't really be any harder than the original question - but things in a class don't always work out like they should.", + "Solution_8": "[quote=\"Kent Merryfield\"]If you were in my class, I would expect you to actually state the answer on #1. :)[/quote]\r\nWhy ? It is the most natural and immediate answer on #1 ! Without l'Hospital or power series which are strong instruments for the \"destruction\" of any limits. Then where is the entertainment ? Or the entertainment is using L'H or power series ..." +} +{ + "Tag": [], + "Problem": "A man is three times as old as his son was at the time when the father was twice as old as his son will be two years from now. Find the present age of each person if the sum of their ages is 55 years.\r\nthanks, it would be nice if you posted an explanation for me. YAY TY PPL", + "Solution_1": "A man is three times as old as his son was at the time when the father was twice as old as his son will be two years from now. Find the present age of each person if the sum of their ages is 55 years. \r\n\r\nThe problem doesn't make sense.", + "Solution_2": "Yeah, it doesnt make sense to me either. Please make sure you word your problems correctly." +} +{ + "Tag": [ + "geometry", + "geometry proposed" + ], + "Problem": "Dear Mathlinkers,\r\nlet ABC a triangle, (1) the incircle of ABC, DEF the contact triangle of ABC, X the second intersection of AD with (1), E' the second intersection of the parallel to AD issue from F with (1), E\" the second intersection of the parallel to BC issue from E' with (1).\r\nProve : E\"X //FD.\r\nSincerely\r\nJean-Louis", + "Solution_1": "\\[ E\"E\\prime \\parallel{}BC \\Longrightarrow DE\\prime(arc) \\equal{} DE\"(arc)\\]\r\n\r\n\\[ FE\\prime\\parallel{}DX \\Longrightarrow DE\\prime(arc) \\equal{} FX(arc)\\]\r\n\r\n\\[ \\Longrightarrow DE\"(arc) \\equal{} FX(arc) Longrightarrow FD\\parallel{}XE\"\\]\r\ni think i didnt understand the question well because obviously it should be harder... :maybe:", + "Solution_2": "Dear stvs, \r\nwhy do you haven't thought at a converse of the Pascal's theorem? \r\nSincerely \r\nJean-Louis" +} +{ + "Tag": [ + "number theory unsolved", + "number theory" + ], + "Problem": "for a positive integer n, let r(n) denote the sum of the remainders when n is divided by 1,2,.....n respectively. prove that r(k)=r(k-1) for infinitely many positive intergers k", + "Solution_1": "$n(mod \\ k)=1+(n-1)(mod \\ k)-\\theta *k$, were $\\theta =1$ if $k|n$ and $\\theta=0$ if $k\\not |n$. It give $r(n)=n-2+r(n-1)-\\sigma(n)+n+1$.\r\nTherefore eqquation $r(n)=r(n-1)$ equavalent to $\\sigma(n)=2n-1$.\r\nIt is obviosly $\\sigma(2^{k})=2^{k+1}-1$. Therefore $r(2^{k})=r(2^{k}-1)$." +} +{ + "Tag": [], + "Problem": "Jennifer, Mike and Carol each have a bunch of quarters. Jennifer and Mike have 26 quarters together. Jennifer and Carol have 20 quarters together. Mike and Carol have 22 quarters together. How many cents does Mike have?", + "Solution_1": "$ J\\plus{}M\\equal{}26$\r\n$ J\\plus{}C\\equal{}20$\r\n$ M\\plus{}C\\equal{}22$\r\n\r\nThree equations and three unknowns. We get $ M\\equal{}14$, $ C\\equal{}8$, and $ J\\equal{}12$. Then Mike has $ 14\\time 25\\equal{}350$ cents.", + "Solution_2": "A quick way of solving this is by adding all equations together we get 2(J+M+C)=68, so J+M+C=34\r\n\r\nSubtracting the second equation from our new one, J+M+C-(J+C)=34-20, so M=14.", + "Solution_3": "The question asked for number of cents, not quarters. I know you could easily get there but if you wrote that, you would be wrong." +} +{ + "Tag": [], + "Problem": "In triangle ABC ,BD and CE are external bisectors of angles B and C. Perpediculars are drawn on BD and CE from A. If the perpenicullars are at ponts Xand Y,prove that XY is parallel to BC.", + "Solution_1": "XY intersects the midpoints of the triangle. Also, let BD CE meet at P. Then XAB XPA are similar etc." +} +{ + "Tag": [ + "calculus", + "integration", + "function", + "complex analysis" + ], + "Problem": "Looking at the following integral:\r\n\r\n$\\int_{0}^{1}\\frac{lnx}{x^{2}+1}$\r\n\r\nSince both $lnx$ and $x^{2}+1$ are analytic at $x_{0}= i$,$i$is a simple pole of the above function which we are integrating, right? And the corresponding residue is $\\frac{ln(i)}{2i}= \\frac{\\pi}{4}.$\r\nIf everything that I have done thus far is correct? How do I use Cauchy's Residue Theorem to finish off the evaluation of the above integral. The case for which either or both of the bounds are at infinity seems easier. I will appreciate any hint.", + "Solution_1": "Perhaps you mean integrating:\r\n\r\n$\\int_{0}^{1}\\frac{ln(x)}{x^{2}+1}dx$\r\n\r\nby considering this integral:\r\n\r\n$\\int_{C}\\frac{Log[z]}{z^{2}+1}dz$\r\n\r\nwhere C is some closed contour indented about the origin and extending past $i$ so as to include the interval (0,1] and one or both of the poles so that Cauchy's Integral Theorem would apply. \r\n\r\nI don't see an easy way to do this however since the upper integration limit conflicts with the pole at $i$ and even if it didn't the resulting integrals may not be easy to evaluate. Brings up an interesting question though: just how close can a different number in the denominator get to 1 and have Cauchy's Integral Theorem still apply if I use a contour with a radius of 1?", + "Solution_2": "Hey, I found this interesting. Try a \"brute-force\" approach using the square contour I plotted below. Now I can say:\r\n\r\n$\\int_{C}\\frac{Log[z]}{z^{2}+1}dz=2\\pi i\\mathop\\text{Res}_{z\\;=\\;i}\\left\\{\\frac{Log[z]}{z^{2}+1}\\right\\}$\r\n\r\nIn that case I can now write:\r\n\r\n$\\int_{0}^{1}\\frac{ln(x)}{x^{2}+1}dx=1/2\\left\\{\\pi i^{2}/2-\\int_{0}^{1}\\frac{\\pi i}{r^{2}+1}dr-I_{2}-I_{3}-I_{4}\\right\\}$\r\n\r\nwhere all the I's are the integrals around contours $C_{1}$ through $C_{6}$. Note I just waved my hands about the part of letting the central circular contour go to zero. Need to to justify that.\r\n\r\nTry doing as much as you can analytically then resort to numerical integration to do the rest. When I do that I get:\r\n\r\n$\\int_{0}^{1}\\frac{ln(x)}{x^{2}+1}dz\\approx-0.915966+i5.32707x10^{-15}$\r\n\r\nPretty close right? Of course I'm not the sharpest knife in here. Maybe someone will suggest an easier way.", + "Solution_3": "According to http://mathworld.wolfram.com/CatalansConstant.html, the answer is $-K$.", + "Solution_4": "Alright, got another one. How about using the second contour below and concluding:\r\n\r\n$\\int_{0}^{1}\\frac{ln(x)}{x^{2}+1}dx=1/2\\left\\{\\frac{\\pi^{2}i}{4}-\\int_{0}^{1}\\frac{\\pi i}{r^{2}+1}dr-\\text{P.V}\\int_{C_{2}}\\frac{Log(z)}{z^{2}+1}dz\\right\\}$\r\n\r\nand just letting not only the contour about the origin go to zero but the one about the pole at $i$ go to zero as well.\r\n\r\nWouldn't that work too?", + "Solution_5": "Hey Shawtoos,\r\nIt seems like you are in the applied track. Those approximations nearly put me off:). I was interested in the closed form of the integral in question. But thank you though. I switched to polar form and hopping to use some nice substitutions, but the quadrature got massier (is massier even an English word?). \r\n\r\nPuuhiki, thank for the link." +} +{ + "Tag": [ + "vector", + "linear algebra", + "matrix", + "linear algebra theorems" + ], + "Problem": "can anybody pleas help me to solve the following questions \r\n :arrow: is it possible that vectors v1,v2,v3, are linearly [u]dependent[/u], but the vectors w1=v1+v2,w2=v2,v3,w3=v3+v1, are linearl [u]independent[/u], explain ur answer\r\n\r\n :arrow: second question is :what the metrix is the zero vector of the space M2X3\r\n\r\nor anybody can recommend me any website to help me solve the questions\r\n\r\nthank you", + "Solution_1": "If $ \\{v_1,v_2,v_3\\}$ is a dependent set, then its span has dimension at most $ 2.$ Then you define $ w_1,w_2,w_3$ in terms of $ v_1,v_2,v_3.$ What you wrote is a little hard to read (there seems to be a typo in it) but it does appear that each of the $ w_j$ is being defined as a linear combination of the $ v_i.$ Hence, they're in that span whose dimension is at most $ 2,$ and there can't be three such independent vectors.\r\n\r\nSo the answer is that what you describe is not possible.\r\n\r\nAs for the second question: if you regard the set of all matrices of a certain size as a vector space, then the zero vector is simply the zero matrix - the matrix whose entries are all zero. \r\n\r\nThe zero element of a vector space has these properties:\r\n\r\nIt's the identity element for addition.\r\n\r\nIf you multiply any member of the vector space by the scalar $ 0,$ you get this zero element.", + "Solution_2": "think you. it makes sense to me now" +} +{ + "Tag": [], + "Problem": "Prove that if p is a prime p/((p-2)!-1). but if p >5 then (p-2)!-1 is not a power of p. :wallbash:", + "Solution_1": "It is immediate from Wilson's Theorem. \r\n[url]http://en.wikipedia.org/wiki/Wilson%27s_Theorem[/url][/url]", + "Solution_2": "Even i did it by wilson's theorem", + "Solution_3": "so wats the fuss all 'bout?" +} +{ + "Tag": [ + "trigonometry" + ], + "Problem": "Find $n\\in\\mathbb{N^{*}}$ such that $(a+ib)^{n}+(b+ai)^{n}$ is real for any $a,b\\in\\mathbb{R^{*}}$.", + "Solution_1": "$n=\\frac{(3 + 4.k).\\pi}{4.x}$ such that $cos\\ x=\\frac{a}{\\sqrt{a^2 + b^2}}$ and $k=\\{0,1,2,3,4,5...\\}$", + "Solution_2": ":? $n$ should have an simple form, not like that.", + "Solution_3": "I got [tex]n=4k,\\ k\\in\\mathbb{Z}^+[/tex]. I'll post the solution a bit later.", + "Solution_4": "Correct. I'm waiting. I hate so much Combinatorics :mad: !", + "Solution_5": "[hide=\"My (long) solution\"]\nIt takes some tedious complex agebra. First we want trigonometric forms for the numbers we are going to sum. Assuming [tex]a,b[/tex] don't simultaneously equal zero we have:\n\n[tex]z_1=a+bi=\\sqrt{a^2+b^2}(\\cos\\alpha +i\\sin\\alpha )[/tex]\n\nwhere [tex]\\alpha =\\arccos\\frac{a}{\\sqrt{a^2+b^2}}=\\arcsin\\frac{b}{\\sqrt{a^2+b^2}}[/tex]\n\n[tex]z_2=b+ai=\\sqrt{a^2+b^2}(\\cos\\beta +i\\sin\\beta )[/tex]\n\nwhere [tex]\\beta =\\arccos\\frac{b}{\\sqrt{a^2+b^2}}=\\arcsin\\frac{a}{\\sqrt{a^2+b^2}}[/tex]\n\nBy de Moivre we get:\n\n[tex]z=z_1^n+z_2^n=(a+bi)^n+(b+ai)^n=\\\\=\\sqrt{a^2+b^2}(\\cos n\\alpha +i\\sin n\\alpha )+\\sqrt{a^2+b^2}(\\cos n\\beta +i\\sin n\\beta )[/tex]\n\nSo we want [tex]\\mbox{Im}z=0[/tex] and that means:\n\n[tex]\\sin n\\alpha +\\sin n\\beta=2\\sin\\frac{n(\\alpha +\\beta )}{2}\\cos\\frac{n(\\alpha -\\beta )}{2}=0[/tex]\n\nBy cyclometric identities we get:\n\n[tex]\\alpha +\\beta =\\arcsin\\frac{b}{\\sqrt{a^2+b^2}}+\\arcsin\\frac{a}{\\sqrt{a^2+b^2}}=\\\\=\\arcsin\\left (\\frac{b}{\\sqrt{a^2+b^2}}\\sqrt{1-\\left (\\frac{a}{\\sqrt{a^2+b^2}}\\right )^2}+\\frac{a}{\\sqrt{a^2+b^2}}\\sqrt{1-\\left (\\frac{b}{\\sqrt{a^2+b^2}}\\right )^2}\\right )=\\\\=\\arcsin\\frac{a^2+b^2}{a^2+b^2}=\\frac{\\pi}{2}[/tex]\n\n[tex]\\alpha - \\beta =\\arccos\\frac{a}{\\sqrt{a^2+b^2}}-\\arccos\\frac{b}{\\sqrt{a^2+b^2}}=\\\\=\\arccos\\left [\\left (\\frac{a}{\\sqrt{a^2+b^2}}\\right )\\left (\\frac{b}{\\sqrt{a^2+b^2}}\\right )+\\sqrt{1-\\left (\\frac{b}{\\sqrt{a^2+b^2}}\\right )^2}\\sqrt{1-\\left (\\frac{a}{\\sqrt{a^2+b^2}}\\right )^2}\\right ]=\\\\=\\arccos\\frac{2ab}{a^2+b^2}[/tex]\n\nFinally it is:\n\n[tex]\\mbox{Im}z=0\\ \\Leftrightarrow\\ \\sin\\frac{n\\pi}{4}=0[/tex]\n\nAnd the result follows.[/hide]", + "Solution_6": "Slizzel, Can you post your solution? Cause I can't think the other solution that different with aappaa's . My solution is the same like aapaa's (first thing that remind me when I see some equation like that is de Moivre). You said that you use combinatoric in your solution :)", + "Solution_7": "\\begin{eqnarray*}\r\nE&=&(a+bi)^n+(b+ia)^n\\\\\r\nE&=&(a+bi)^n+i^n(a-bi)^n\\\\\r\nE&=&a^n+\\binom{1}{n}a^{n-1}bi+\\binom{2}{n}a^{n-2}b^2i^2+...+\\binom{n}{n}b^ni^n+i^n\\left(a^n-\\binom{1}{n}a^{n-1}bi+\\binom{2}{n}a^{n-2}b^2i^2-...+(-1)^n\\binom{n}{n}b^ni^n\\right)\r\n\\end{eqnarray*}\r\n\r\nTo have $E\\in\\mathbb{R}$ then $i^n=1$ so $n=4k$ for any integer $k$", + "Solution_8": "Oh...I forgot the binomial expansion :P" +} +{ + "Tag": [ + "limit", + "calculus", + "calculus computations" + ], + "Problem": "Evaluate $ \\lim_{x \\to \\infty} x\\left((x^2 \\plus{} 1)^{\\frac{1}{2}} \\minus{} (x^3 \\plus{} 1)^{\\frac{1}{3}}\\right)$", + "Solution_1": "$ \\begin{array}{l}\r\n \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\left[ {x\\left( {\\sqrt {{x^2} \\plus{} 1} \\minus{} \\sqrt[3]{{{x^3} \\plus{} 1}}} \\right)} \\right] \\equal{} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\left[ {x\\left( {\\sqrt {{x^2} \\plus{} 1} \\minus{} \\sqrt {{x^2}} \\minus{} \\sqrt[3]{{{x^3} \\plus{} 1}} \\plus{} \\sqrt[3]{{{x^3}}}} \\right)} \\right] \\equal{} \\\\ \r\n \\equal{} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\left[ {x\\left( {\\frac{1}{{x\\left( {\\sqrt {1 \\plus{} \\frac{1}{{{x^2}}}} \\plus{} 1} \\right)}} \\minus{} \\frac{1}{{{x^2}\\left( {\\sqrt[3]{{1 \\plus{} \\frac{2}{{{x^3}}} \\plus{} \\frac{1}{{{x^6}}}}} \\plus{} \\sqrt[3]{{1 \\plus{} \\frac{1}{{{x^3}}}}} \\plus{} 1} \\right)}}} \\right)} \\right] \\equal{} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\left( {\\frac{1}{{\\left( {\\sqrt {1 \\plus{} \\frac{1}{{{x^2}}}} \\plus{} 1} \\right)}} \\minus{} \\frac{1}{{x\\left( {\\sqrt[3]{{1 \\plus{} \\frac{2}{{{x^3}}} \\plus{} \\frac{1}{{{x^6}}}}} \\plus{} \\sqrt[3]{{1 \\plus{} \\frac{1}{{{x^3}}}}} \\plus{} 1} \\right)}}} \\right) \\equal{} \\frac{1}{2} \\\\ \r\n \\end{array}$", + "Solution_2": "That is correct.", + "Solution_3": "I would have used the binomial series:\r\n\r\n$ (x^2\\plus{}1)^{1/2}\\equal{}x\\left(1\\plus{}\\frac1{x^2}\\right)^{\\frac12}$\r\n\r\n$ \\equal{}x\\left(1\\plus{}\\frac1{2x^2}\\plus{}O(x^{\\minus{}4})\\right)\\equal{}x\\plus{}\\frac1{2x}\\plus{}O(x^{\\minus{}3}).$\r\n\r\n$ (x^3\\plus{}1)^{1/3}\\equal{}x\\left(1\\plus{}\\frac1{x^3}\\right)^{\\frac13}$\r\n\r\n$ \\equal{}x\\left(1\\plus{}\\frac1{3x^3}\\plus{}O(x^{\\minus{}6})\\right)\\equal{}x\\plus{}\\frac1{3x^2}\\plus{}O(x^{\\minus{}5}).$\r\n\r\nSo $ x\\left((x^2\\plus{}1)^{\\frac12}\\minus{}(x^3\\plus{}1)^{\\frac13}\\right)\\equal{}x\\left(x\\plus{}\\frac1{2x}\\plus{}O(x^{\\minus{}3})\\minus{}x\\minus{}\\frac1{3x^2}\\plus{}O(x^{\\minus{}5})\\right)$\r\n\r\n$ \\equal{}\\frac12\\plus{}O(x^{\\minus{}1}).$\r\n\r\nSo the limit is $ \\frac12.$", + "Solution_4": "Yes, the binomial expansion is a powerful tool for problems like this. The shocking truth is that a considerable number of calculus students rarely appeal to this powerful tool." +} +{ + "Tag": [ + "function", + "algebra proposed", + "algebra" + ], + "Problem": "Determine all functions$ f : N \\rightarrow N$ such that for all $ n\\epsilon N$\r\nf(n) + f(n + 1) = f(n + 2)f(n + 3) - 1996:", + "Solution_1": "first take the case where f(n) = 1 for some n > 2.\r\nThen starting from f(n-2) the sequence goes, a,b,1,a+b,(b+1)/(a+b)\r\nbut this must be in N so a = 1, then f(n+2) = b+1.\r\nIn this case the sequence is a subsequence of \r\n\r\n1,1,1,2,1,3,1,4,1,5,1,6,...\r\n\r\nsecond assume that f(n) > 1 for all n > 2, but f(n) = 2, for some n > 2\r\nthen starting from f(n-2) the sequnce goes a,b,2,(a+b)/2, (2b+4)/(a+b)\r\nbut (2b+4)/(a+b) must be > 1 => a = 2, but then we find that the sequence must be\r\n\r\n2,2,2,2,2,.....\r\n\r\nfinally consider the case where f(n) > 2 for all n > 2\r\nthen the sequence proceeds a,b,c,(a+b)/c\r\nbut a+b > (a+b)/c + c, if both c and (a+b)/c are > 2\r\nthis means the sum of alternate pairs forms a decreasing sequence, contradiction, so there are no other possibilities" +} +{ + "Tag": [ + "quadratics", + "function", + "modular arithmetic", + "algebra", + "polynomial", + "Euler", + "number theory" + ], + "Problem": "Hi,\r\n I have been reading the \"An Introduction to the theory of Numbers' by Hardy & Wright. In this in the section explaining \"quadratic residues\", I am facing some difficulty and I hope someone can help me.\r\n\r\nit has been proved/arrived at that.. a^((p-1)/2) = (a/p) (mod p) where (a/p) is the Legendre's symbol. If I start from Fermat's Little theorem a^(p-1) = 1 (mod p) and take sqaure root, I get abs{ (a^((p-1)/2)) } = 1 (mod p)... where abs{} is the absolute function\r\n\r\nnow If I assume (a/p) = 1, then there exists some x such that x^2 = a (mod p). now if I raise a to power of (p-1)/2, then I get the solution for a^((p-1)/2) = 1 (mod p). Here it is same as (a/p) when a is a quadratic residue mod p. But I am not able to reason for -1? Can anyone help me?\r\n\r\nAlso, it has also been stated that...\r\n(p-1)! = -(a/p)a^((p-1)/2) (mod p). If possible, please provide proof for this also or direct me to the place where I can find these two proofs?\r\n\r\nAny help will be highly appreciated,\r\n\r\nThanks in advance.\r\n\r\nregards\r\nArvind", + "Solution_1": "[quote=\"ArvindS\"]it has been proved/arrived at that.. a^((p-1)/2) = (a/p) (mod p) where (a/p) is the Legendre's symbol. If I start from Fermat's Little theorem a^(p-1) = 1 (mod p) and take sqaure root, I get abs{ (a^((p-1)/2)) } = 1 (mod p)... where abs{} is the absolute function\n\nnow If I assume (a/p) = 1, then there exists some x such that x^2 = a (mod p). now if I raise a to power of (p-1)/2, then I get the solution for a^((p-1)/2) = 1 (mod p). Here it is same as (a/p) when a is a quadratic residue mod p. But I am not able to reason for -1? Can anyone help me?\n[/quote]\n\nLet me rephrase your words in a clearer manner.\n\n[u][b]Theorem[/b][/u]\nLet $p$ be an odd prime and $a$ be an integer.\nIf $a$ is a quadratic residue modulo $p$ then $a^{\\frac{p-1}{2}}\\equiv 1\\pmod{p}$ (1)\nIf $a$ is a quadratic non residue modulo $p$ then $a^{\\frac{p-1}{2}}\\equiv -1\\pmod{p}$ (2)\n\n[u]Proof[/u]\nBy Fermat's Little Theorem $a^{p-1}\\equiv 1\\pmod{p}$. Therefore we have\n\\[ \\left( a^{\\frac{p-1}{2}}-1\\right)\\left( a^{\\frac{p-1}{2}}+1\\right)\\equiv 0\\pmod{p} \\]\nOne and only one of these two factors is divisible by $p$. So one and only one of the congruences (1) and (2) holds.\nIf $a$ is a quadratic residue then there is a $b$ such that $a\\equiv b^2\\pmod{p}$. In this case\n\\[ a^{\\frac{p-1}{2}}\\equiv b^{p-1}\\equiv 1\\pmod{p} \\].\nSo $a$ is a solution to (1). Since (1) is a polynomial equation with degree $\\frac{p-1}{2}$, by Lagrange's Theorem, it cannot have more than $\\frac{p-1}{2}$ solutions. Hence the quadratic residue are exactly the solutions to (1). It then also follows that the quadratic non residues are not solutions to (1) and so they satisfy (2).\n\n[u]Corollary: Euler Criterion[/u]\n$(\\frac{a}{p})\\equiv a^{\\frac{p-1}{2}}\\pmod{p}$\n\n[quote=\"ArvindS\"]Also, it has also been stated that...\n(p-1)! = -(a/p)a^((p-1)/2) (mod p). If possible, please provide proof for this also or direct me to the place where I can find these two proofs?\n[/quote]\r\nI think $p$ here is a prime.\r\nThis is just a direct result from Wilson's Theorem.\r\n$(p-1)!\\equiv -1\\pmod{p}$\r\nand also the formula $(\\frac{a}{p})^2=1$" +} +{ + "Tag": [ + "geometry", + "circumcircle", + "incenter", + "geometry unsolved" + ], + "Problem": "Given a triangle ABC. Let O be the circumcenter of this trangle ABC. The incircle of triangle ABC has center I. Let D,E,F be the feet of the altitudes of triangle ABC from the vertices A,B,C, respectively. Prove that the three points E,I,F are collinear if and only if the three points D,I,O are collinear.", + "Solution_1": "[quote=\"Fang-jh\"]Given a triangle ABC. Let O be the circumcenter of this trangle ABC. The incircle of triangle ABC has center I. Let D,E,F be the feet of the altitudes of triangle ABC from the vertices A,B,C, respectively. Prove that the three points E,I,F are collinear if and only if the three points D,I,O are collinear.[/quote\r\nWe need [b]4 lemma [/b] for this sum :)\r\n(This 's in IMO SHortlisted )", + "Solution_2": "Let $\\alpha,\\beta$ and $\\gamma$ (assume $\\beta \\geq \\gamma$) be angles of triangle $ABC$ and let $CI,AI$ intersect $EF$ at $L,M$ respectively.Then \r\n$\\frac{EL}{LF}=\\frac{cos \\gamma sin\\frac{\\gamma}{2}}{cos(\\frac{\\gamma}{2}+\\alpha)sin \\beta}$ and\r\n${\\frac{EM}{MF}=\\frac{AE}{AF}}=\\frac{sin \\gamma }{sin \\beta}$\r\nThen $E,I,F$ collinear iff $\\frac{cos \\gamma sin\\frac{\\gamma}{2}}{cos(\\frac{\\gamma}{2}+\\alpha)sin \\beta}=\\frac{sin \\gamma }{sin \\beta}$\r\n$cos \\gamma=2cos(\\frac{\\gamma}{2}+\\alpha)cos \\frac{\\gamma}{2}$\r\n$cos \\gamma+cos \\beta=cos \\alpha$\r\nLet $S$ be foot of perpendicular from $O$ to $CB$ and let $R,r$ denotes circumcenter and incenter,respectively.\r\nThen $O,I,D$ collinear iff $\\frac{r}{OS}=\\frac{p-b-BD}{\\frac{a}{2}-BD}$\r\n$\\frac{r}{OS}=\\frac{a+c-b-2BD}{a-2BD}=\\frac{a+c-b-\\frac{a^{2}+c^{2}-b^{2}}{a}}{a-\\frac{a^{2}+c^{2}-b^{2}}{a}}=\\frac{c+b-a}{c+b}=\\frac{2(p-a)}{c+b}$\r\n$\\frac{(p-a)tan\\frac{\\alpha}{2}}{R.cos\\alpha}=\\frac{2(p-a)}{2R(sin \\gamma+sin\\beta)}$\r\n${tan\\frac{\\alpha}{2}}(sin \\gamma+sin\\beta)=cos\\alpha$\r\n${2sin(\\frac{\\gamma+\\beta}{2})cos(\\frac{\\gamma-\\beta}{2})tan\\frac{\\alpha}{2}}=cos\\alpha$\r\n$2sin\\frac{\\alpha}{2}cos(\\frac{\\gamma-\\beta}{2})= cos \\alpha$\r\n$cos \\gamma+cos \\beta=cos \\alpha$\r\nSo each collinearity holds iff $cos \\gamma+cos \\beta=cos \\alpha$" +} +{ + "Tag": [ + "trigonometry", + "function", + "calculus", + "integration" + ], + "Problem": "The period of $ x$ is $ \\frac{1}{2}$. The period of $ y$ is $ \\frac{2}{5}$. $ x$ and $ y$ are both trig functions. If $ x$ and $ y$ are added, what is the period of the resulting function?\r\n\r\nThanks in advance!", + "Solution_1": "The period would be the smallest number that is an integer multiple of both numbers, which is $ 2$.", + "Solution_2": "All you actually know is that the period \"divides\" $ 2$. The problem, as stated, doesn't give enough information (depending on exactly what is meant by \"trig function\") to uniquely determine the period.", + "Solution_3": "Oh yeah, good point. I was assuming he meant the simplest case possible, which isn't always the right thing to do.", + "Solution_4": "Actually, I was looking for the simplest case. How do you know that each period \"divides\" 2? What does the quotes mean? Thanks!", + "Solution_5": "What I mean is that you know that the function is periodic with period $ 2$; what you don't know is whether this is the [b]minimal[/b] period. [i]A priori[/i], the period could be any rational number $ q$ such that $ \\frac{2}{q}$ is an integer, which is what I meant by \"divides\"; it's not divisibility in the strict sense, but it's the relevant notion for periodic functions.", + "Solution_6": "Ok, Thanks a lot! :)", + "Solution_7": "The period of each graph is going to go through an integral number of periods in one period of their sum. In essence, you're looking for the smallest $ (p,q)$ such that $ \\frac{2p}{5}\\equal{}\\frac{q}{2}$. This will happen when $ (p,q)\\equal{}(5,4)$, which gives us a period of $ 2$." +} +{ + "Tag": [ + "geometry", + "3D geometry", + "sphere", + "advanced fields", + "advanced fields unsolved" + ], + "Problem": "I have already asked the question before but I didnot get an answer.\r\n For what n, $P^{n}(R)$ orientable?\r\n My guess is for odd n.\r\n Because $P^{2}$ contains a mobius strip.(I am trying to show it but going nowhere)", + "Solution_1": "The sphere $S^{n}\\subset\\mathbb R^{n+1}$ is orientable, and $\\mathbb RP^{n}$ is obtained from it by identifying the antipodal points. When $n$ is odd, the antipodal map preserves the orientation of the sphere, which makes for an orientation on $\\mathbb RP^{n}$. When $n$ is even, the antipodal map reverses orientation, and this makes $\\mathbb RP^{n}$ nonorientable", + "Solution_2": "Thank you very much mlok. \r\nwhat about the answer to my next question?", + "Solution_3": "You didn't really state another question. Maybe you accidentally deleted part of it.", + "Solution_4": "Yes I did,I asked how I can 'see' $P^{2}$ contains a mobius strip.", + "Solution_5": "The [url=http://en.wikipedia.org/wiki/Real_projective_plane]projective plane[/url] can be realized as a square with opposite sides glued in a weird way. If you make a cut where the vertical sides were glued together, you get a Mobius strip. Making a cut like this is topologically equivalent to cutting out a hole." +} +{ + "Tag": [ + "trigonometry", + "logarithms" + ], + "Problem": "I have a math placement test in a few days and I needed some help with some of these problems!\r\nI was working on these sample problems provided by the school site and needed some help\r\n\r\nhttp://www.math.uic.edu/undergrad/mathplacementtest.pdf\r\n\r\nI can't figure out \r\n#11\r\n#13\r\n#14\r\n#15\r\nCan someone walk me through these problems? THank you very much.", + "Solution_1": "[quote=\"nitric0\"]I have a math placement test in a few days and I needed some help with some of these problems!\nI was working on these sample problems provided by the school site and needed some help\n\nhttp://www.math.uic.edu/undergrad/mathplacementtest.pdf\n\nI can't figure out \n#11\n#13\n#14\n#15\nCan someone walk me through these problems? THank you very much.[/quote]\r\n\r\n:)\r\n#11: use the log rules: $\\log(AB) = \\log A + \\log B$, $\\log(A^b)=b\\log A$\r\n\r\n#13: draw a triangle! \r\n\r\n#14: use the trigonometric relation $\\sin^2 x + \\cos^2 x=1$\r\n\r\nand #15: work with the second half first: ($\\cot x + \\tan x$).\r\n\r\nGood luck on the test!", + "Solution_2": "[hide=\"Answers\"]\n#11: E\n#13: D\n#14: C\n#15: D\n[/hide]\r\nIf u still didn't get it, I can post the solutions too\r\nGood Luck", + "Solution_3": "Is there anyway you guys can walk me through these? I totally forgot how to do them, and I have one more problem to ask! (#9)", + "Solution_4": "These are very straightforward using the hints that cincodemayo5590 provided, but I'll post the solutions to a couple anyways if you want.\r\n\r\n[hide=\"Problem 9\"]$\\frac{\\sqrt{3x+2}}{\\left(3x+2\\right)^{\\frac{7}{2}}}=\\left(3x+2\\right)^{\\frac{1}{2}-\\frac{7}{2}}=\\left(3x+2\\right)^{-3}\\implies\\boxed{B}$[/hide]\n\n[hide=\"Problem 15\"]$\\cos^{2}x\\left(\\tan x +\\cot x\\right)=\\cos^{2}x\\left(\\frac{\\tan^{2}x+1}{\\tan x}\\right)=\\cos^{2}x\\left(\\frac{\\sec^{2}x}{\\tan x}\\right)=\\frac{1}{\\tan x}=\\cot x\\implies\\boxed{D}$[/hide]", + "Solution_5": "Wow thanks amcavoy, that really helped me. Can you show me the other problems so I'm set? I tried them again and am struggling with them once again. :(", + "Solution_6": "Why don't you post what you've done, and we can help you with your specific problems?", + "Solution_7": "[hide=\"Problem 11\"]$\\log\\frac{8A^\\frac{1}{4}}{C^{3}}=\\log 8A^{\\frac{1}{4}}-\\log C^{3}=\\log 8 +\\frac{1}{4}\\log A-3\\log C\\implies\\boxed{E}$[/hide]\n\n[hide=\"Problem 13\"]You know that $\\sin^{2}x+\\cos^{2}x=1$, so you have\n\n\\[ \\tan x=\\frac{\\sin x}{\\cos x}=\\frac{\\sqrt{1-\\cos^{2}x}}{\\cos x}=\\frac{3\\sqrt{1-\\frac{a^{2}}{9}}}{a}=\\frac{\\sqrt{9-a^{2}}}{a}\\implies\\boxed{D} \\][/hide]\n\n[hide=\"Problem 14\"]$\\sin^{2}x+\\cos x=1\\implies 1-\\cos^{2}x+\\cos x=1\\implies\\cos x\\left(\\cos x-1\\right)=0$\n\nSo $\\cos x=0$ when $x=\\frac{\\pi}{2},\\,\\frac{3\\pi}{2}$ and $\\cos x=1$ when $x=0$, so the answer is $\\boxed{C}$.[/hide]\r\n\r\nThere you go." +} +{ + "Tag": [ + "geometry", + "circumcircle", + "Olimpiada de matematicas" + ], + "Problem": "Sea M el punto medio de la mediana AD del triangulo ABC(D pertenece al lado BC) la recta BM corta al lado AC en el punto N. Demuestre que AB es tangente a la circunferencia circunscrita al triangulo NBC si, y solamente si, se verifica $BM/MN=BC^{2}/BN^{2}$", + "Solution_1": "[color=darkred][b]Enunciation.[/b]Given is a triangle $ABC$ . Denote the middlepoints $D,M$ of the segments $[BC]$ , $[AM]$ and the point $N\\in BM\\cap AC$ . \nProve that the line $AB$ is tangent to the circumcircle of the triangle $BCN$ $\\Longleftrightarrow$ $\\frac{MB}{MN}=\\left(\\frac{BC}{BN}\\right)^{2}$ $\\Longleftrightarrow$ $AC=AB\\sqrt 3\\ .$[/color]\r\n\r\n[color=darkblue][b]Proof.[/b] Aplly the Menelaus' theorem to the transversal $\\overline{BMN}$ and the triangle $ADC\\ :$ $\\frac{BD}{BC}\\cdot\\frac{NC}{NA}\\cdot \\frac{MA}{MD}=1$ $\\Longleftrightarrow$ $\\frac{NC}{2}=\\frac{NA}{1}=\\frac{b}{3}\\ .$\n\nApply the Menelaus' theorem to the transversal $\\overline{AMD}$ and the triangle $BNC\\ :$ $\\frac{AN}{AC}\\cdot\\frac{DC}{DB}\\cdot\\frac{MB}{MN}=1$ $\\Longleftrightarrow$ $\\frac{MB}{3}=\\frac{MN}{1}=\\frac{BN}{4}\\ .$\n\nApply the teorem of the $B$- median in the triangle $ABD\\ :$ $4\\cdot BM^{2}=2\\left(c^{2}+\\frac{a^{2}}{4}\\right)-m^{2}_{a}$ \n\nand $4m^{2}_{a}=2(b^{2}+c^{2})-a^{2}$ $\\Longrightarrow$ $16\\cdot BM^{2}=3a^{2}-2b^{2}+6c^{2}$ $\\Longrightarrow$ $9\\cdot BN^{2}=3a^{2}-2b^{2}+6c^{2}\\ .$\n\n$1.\\blacktriangleright$ $\\frac{MB}{MN}=\\left(\\frac{BC}{BN}\\right)^{2}$ $\\Longleftrightarrow$ $\\sqrt 3=\\frac{a}{BN}$ $\\Longleftrightarrow$ $BN=\\frac{a\\sqrt 3}{3}$ $\\Longleftrightarrow$ $3a^{2}-2b^{2}+6c^{2}=3a^{2}$ $\\Longleftrightarrow$ $b=c\\sqrt 3\\ .$\n\n$2.\\blacktriangleright$ The line $AB$ is tangent to the circumcircle of the triangle $NBC$ $\\Longleftrightarrow$ $\\triangle ABN\\sim\\triangle ACB$ $\\Longleftrightarrow$ $\\frac{c}{b}=\\frac{BN}{a}=\\frac{NA}{c}$ $\\Longleftrightarrow$ \n\n$BN=\\frac{ac}{b}$ and $NA=\\frac{c^{2}}{b}$ $\\Longleftrightarrow$ $BN=\\frac{ac}{b}$ and $b=c\\sqrt 3$, i.e. $BN=\\frac{a\\sqrt 3}{3}$ $\\Longleftrightarrow$ $b=c\\sqrt 3\\ .$[/color]", + "Solution_2": "Es bien sabido que la recta $ BM$ divide a $ AC$ en dos partes siendo una el doble de la otra $ \\Longrightarrow$ $ CN \\equal{} 2AN.$ Por consiguiente al aplicar teorema de Menelao en el $ \\triangle ADC$ cortado por la transversal $ \\overline{BMN}$ se obtiene $ \\frac {_{BM}}{^{MN}} \\equal{} 3.$ \n\nHaciendo uso del Teorema de Stewart en la ceviana $ BN$ del $ \\triangle ABC$ resulta\n\n$ AC \\cdot BN^2 \\equal{} AN \\cdot BC^2 \\plus{} CN \\cdot BA^2 \\minus{} AN \\cdot CN \\cdot AC$ \n\n$ \\Longrightarrow \\ BN^2 \\equal{} \\frac {_1}{^3}BC^2 \\plus{} \\frac {_2}{^3}BA^2 \\minus{} \\frac {_2}{^9}AC^2 \\Longrightarrow \\ \\frac {BC^2}{BN^2} \\equal{} \\frac {BC^2}{\\frac {1}{3}BC^2 \\plus{} \\frac {2}{3}BA^2 \\minus{} \\frac {2}{9}AC^2}$\n\n$ \\frac {BM}{MN} \\equal{} \\frac {BC^2}{BN^2} \\Longleftrightarrow \\ \\frac {BC^2}{BN^2} \\equal{} \\frac {BC^2}{\\frac {1}{3}BC^2 \\plus{} \\frac {2}{3}BA^2 \\minus{} \\frac {2}{9}AC^2} \\equal{} 3$ \n\n$\\Longleftrightarrow \\ BA^2 \\equal{} \\frac {_1}{^3}AC^2 \\equal{} AN \\cdot AC \\Longleftrightarrow$ $ AB$ es tangente a la circunferencia $ \\odot(NBC).$" +} +{ + "Tag": [ + "articles", + "calculus", + "AMC", + "USA(J)MO", + "USAMO", + "AIME" + ], + "Problem": "I recently became very interested in the development of brain after reading an article about a youngster who completed high school maths in his first grade and started college lvl calculus in his 2nd grade.\r\n\r\nAccording to his father, he followed a EQ program so-called the \"prodigy program\" (sorry/// I didnt know how to translate this! Its a korean word)\r\nwhich included several activities that made more frequent use of his left (its either left or right... its the part which most ppl dont use very often) brain. \r\n\r\nAnd according to experts, our brain rapidly develops from age 1 to 4 and the rate of growth declines around at age of 6. \r\n\r\nAlthough in Olympiads or USAMO's there must be students who have worked exceptionally hard, I personally think that the vast majority of them are those who made more frequent use of the left side of the brain when they were at a very young age (or, play with puzzles/legos to develop their ability to think mathematically/imaginatively)\r\n\r\nDo you also think that most of your mathematical abilities judged upon primary factors than secondary? (The word secondary here means effort and hard work)\r\n\r\nOf course, you do have to put some effort in but dont you agree?\r\n\r\nI've read a post where this USAMO person who studied pre-calc in his senior year. \r\n\r\nSo if I sum it up:\r\n- You can still develop your mathematical skills enormously thru solving questions and putting effort, however, there is an extent; I most certainly do not think \"ordinary\" people can achieve a gold in Olympiads without a early stage of brain development. So its mathematical abilities are much more primary than secondary when it comes to olympiad questions\r\n\r\n\r\n\r\n\r\nPS. Sorry for the occasional (or frequent) grammatical erros. As english isnt my first language, I tend to make mistakes very often. And I also apologise for the lack in fluency in the post. I sound rather arrogant.. Please forgive me;", + "Solution_1": "Hm? I don't think you sounded arrogant at all, and your english seemed pretty good to me, unless I just missed all your errors.\r\n\r\nAnyway, I think that it's kind of the nature or nurture question with genetics. When I was young, I often played with legos, did math, and other kind of stuff like that. I'm not saying that I'm really good at math, which I'm not compared to most people on this forum. My opinion is similar to that of the person you quoted, that much of it has to do with how hard you work and practice, but becoming excellent in math contests is easier depending on your brain development in first several years of one's life.", + "Solution_2": "I feel that I did a good job in high school working from being someone who wouldn't ordinarily be considered very good at math (at least by people on this forum) to one who would. In other words, I don't really think I have much talent. I was (and still am) completely fascinated by mathematics and have always wanted to learn as much of it as possible. It didn't hurt that people were beating me in competitions, and I wanted that to stop as soon as possible. I also perhaps benefited from being excessively stubborn and rebellious, generally in a good way. My mother said so frequently that Asian people tend to be smart, and, being a rational person, I didn't understand why they should be smarter than the rest of us. So I spent a good deal of effort demonstrating that I could do as well as the best of them. Well, that's my ramble for now.", + "Solution_3": "I'm an Asian (Korean) and yet I still struggle in maths Lol!\r\n\r\nBut I onli lived in Korea for like 3 years in total I think and I was born in the US. \r\n\r\nI posted this question on one of the korean forums I go to and many have replied that 90% is a secondary factor. \r\n\r\nI still firmly believe that first few years of brain development are crucial, but here is a new question:\r\n\r\nDo you think that so-called \"ordinary\" people like me can eventaully solve some USAMO questions after overcoming tremendous amount of work?\r\n\r\nOr are ordinary brains limited to AIME questions...\r\n\r\nAs the description for USAMO says, it requires extensive creativity and inventiveness (lol I'm not sure whether that word exists or not). Today I was on the bus imagining diagrams to prove why sum of exterior angles add up to 360 (is this rite anyways? :P) and I just couldnt grasp a vivid clear image. It kept on disappearing and fading away even thou it was a very simple diagram of a hexagon with lines drawn on it. \r\n\r\nHuman Evolution is a extraordinarily fascinating thing.", + "Solution_4": "I am half korean. Do you have a website that has this information?", + "Solution_5": "http://monthly.chosun.com/board/view_content.asp?tnu=200405100053&catecode=&cPage=1\r\n\r\nThis is the original article about the kid I was talking about...\r\n\r\nAnd I visited websites that had info. about some of the techniques the father applied to his genius son. Like Calvitae and some others. \r\n\r\nSorry for the non-koreans... I didnt know how to translate any of those stuff on it.", + "Solution_6": "While primary skills might help, people who did constant puzzles, logic games, and math as young children need to 'keep up the good work' to be cliched. I have friends who were very good at math until the age of eleven or so when they stopped working at it. A bit of natural ability (or maybe just some determination) is needed, but it's the hard work that ultimately decides it." +} +{ + "Tag": [], + "Problem": "Open for discussion.", + "Solution_1": "it looks like a lame copy of narnia", + "Solution_2": "We read the book in 6th grade. I thought it was okay; not the best assigned reading we had but not the worst either.\r\n\r\nKnowing the book I would doubt that the movie would be a copy of Narnia (the book is not a fantasy novel), but I haven't seen the movie before so I can't make any definite conclusions.", + "Solution_3": "It was the worst book i ever read. It was pure torture.", + "Solution_4": "[quote=\"math92\"]It was the worst book i ever read. It was pure torture.[/quote]\r\nHave you ever read A Tale of Two Cities?", + "Solution_5": "Have you read Great Expectations?", + "Solution_6": "have you ever read last of the mohicans?", + "Solution_7": "I read this book in 5th grade.\r\n\r\nI must be honest.\r\n[hide]It was purely the worst book I have ever read in school. Although No Promises in the Wind comes close.\n\n\nGaaahhh![/hide]", + "Solution_8": "Read it in class and saw the original movie in 3rd grade.\r\nI don't remember anything about it, except that the girl dies i think.", + "Solution_9": "We used it as read-aloud in 4th grade. It was sooo sad...but now, I just find it lame.", + "Solution_10": "*feathers start becoming stained black, and eyes start glowing*\r\n*calms down and turns to normal form*\r\nPoint 1: It is not a copy of Narnia. It tells about a girl and her friend who are inspired by the stories.\r\nPoint 2: While it's a bit boring at times, the parts in Terabithia are really good.\r\nPoint 3: BtT is not trash. Let anyone who thinks different face me." +} +{ + "Tag": [ + "analytic geometry", + "graphing lines", + "slope", + "geometry", + "trapezoid" + ], + "Problem": "$20)$ In the coordinate plane, an L-shaped region is bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1), (5,0).$ Find the slope of the line passing through the origin that divides the area of the region in half.", + "Solution_1": "[quote=\"now a ranger\"]$20)$ In the coordinate plane, an L-shaped region is bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1), (5,0).$ Find the slope of the line passing through the origin that divides the area of the region in half.[/quote]\r\n[hide]Since the shape we create after slicing the area in half will be a trapezoid (kind of obvious after you draw it...), it's area will be $\\frac{3(3+x)}{2}$. The area or the original L-shape is 11 (figured out easily). $\\frac{11}{2}=\\frac{3(3+x)}{2}$. Then $11=3(3+x)=9+3x \\rightarrow 3x=2 \\rightarrow x=\\frac{2}{3}$. Where the line that divides the \"L\" into 2 equal-area shapes and the side of the L intersect is $(3, \\frac{7}{3})$.\nTherefore, the slope is $\\frac{\\frac{7}{3}}{3}$ or $\\frac{7}{9}$ :P [/hide]" +} +{ + "Tag": [], + "Problem": "Friends Megan and Heather go to different schools. Megan has math class during first period on each of the 180 days she goes to school. She will be in math class a total of 8640 minutes this year. Heather\u2019s school year also has 180 days, but her math class only meets every other day, so her class periods are longer. (One day she\u2019ll have math class, and then the next day she won\u2019t.) If Megan and Heather end up with the same number of minutes of math class each year, how long are the class periods at Heather\u2019s school?", + "Solution_1": "[hide]8640/180 = 48 minutes per day. Heather is twice that or 96 minutes per day. [/hide]", + "Solution_2": "[quote=\"math92\"]Friends Megan and Heather go to different schools. Megan has math class during first period on each of the 180 days she goes to school. She will be in math class a total of 8640 minutes this year. Heather\u2019s school year also has 180 days, but her math class only meets every other day, so her class periods are longer. (One day she\u2019ll have math class, and then the next day she won\u2019t.) If Megan and Heather end up with the same number of minutes of math class each year, how long are the class periods at Heather\u2019s school?[/quote]\r\n\r\n[hide]Megan's math class lasts $\\frac{8640}{180}=48$ minutes each.\n\nHeather's classes last twice as long, so they last $96$ minutes each.[/hide]", + "Solution_3": "[quote=\"math92\"]Friends Megan and Heather go to different schools. Megan has math class during first period on each of the 180 days she goes to school. She will be in math class a total of 8640 minutes this year. Heather\u2019s school year also has 180 days, but her math class only meets every other day, so her class periods are longer. (One day she\u2019ll have math class, and then the next day she won\u2019t.) If Megan and Heather end up with the same number of minutes of math class each year, how long are the class periods at Heather\u2019s school?[/quote]\r\n[hide]\n8640/180=48 minutes each period for Megan.. Heather has math half of the year, so her periods are twice as long... 48 x 2= 96 minutes[/hide]", + "Solution_4": "[quote=\"math92\"]Friends Megan and Heather go to different schools. Megan has math class during first period on each of the 180 days she goes to school. She will be in math class a total of 8640 minutes this year. Heather\u2019s school year also has 180 days, but her math class only meets every other day, so her class periods are longer. (One day she\u2019ll have math class, and then the next day she won\u2019t.) If Megan and Heather end up with the same number of minutes of math class each year, how long are the class periods at Heather\u2019s school?[/quote]\r\n[hide]$8640/(180/2)= 8640/90= 96$ minutes or 1 hour and 36 minutes.[/hide]" +} +{ + "Tag": [ + "USAMTS", + "trigonometry", + "function", + "geometry", + "logarithms", + "conics", + "parabola" + ], + "Problem": "[url=http://www.unl.edu/amc/e-exams/e6-amc12/factoids12.shtml]Rules[/url]\r\n[b][u]1. 75 minutes[/u][/b]\r\n2. All calculators allowed on AMC are allowed\r\n3. 1.5 per unanswered question, 6 points per correct question\r\n4. Answers are due by [b]8:00 PM EST January 10, 2006[/b] (so all you USAMTS slackers have an extra couple days :P )\r\n\r\nThanks to [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=1846]beta[/url] for checking the questions!\r\n\r\nSubmitting answers: PM me all of your answers! I'd also like to see what you thought about the test, your thoughts on various problems, etc.\r\n\r\nNote: DEFINITION OF PROJECTION: If we have a point X and a plane Y and we draw line $l$ through X and perpendicular to Y, then the point that is the intersection of line $l$ and plane Y is the projection of X onto Y.\r\n\r\nDefinition of Argument: The argument of a complex number z is the angle of that complex number, i.e. $\\arg{\\left(a+bi\\right)}= \\tan^{-1}{\\frac{b}{a}}$.\r\n\r\n\t[hide]1. What is $1+1$?\n\\[\\text{(A) }2\\qquad \\text{(B) }4\\qquad \\text{(C) }67\\qquad \\text{(D) }\\text{limacon}\\qquad \\text{(E) }\\text{I'm clueless!}\\]\n2. If $f(x)$ is an odd function and $g(x)$ is an even function, which of the following functions must be odd?\n\nI. $f(x) \\cdot g(x)$\n\nII. $f(x)+g(x)$\n\nIII. $f(x)-g(x)$\n\\[\\text{(A) }I \\qquad \\text{(B) }II \\qquad \\text{(C) }II,III \\qquad \\text{(D) }I,III \\qquad \\text{(E) }I,II,III \\]\n3. What is the area bounded by the lines $x=1$, $x=2$, $y=0$, and $y=mx+b$ if $m+b=4$ and $2m+b=7$?\n\\[\\text{(A) }5\\qquad \\text{(B) }5.5\\qquad \\text{(C) }6\\qquad \\text{(D) }7.5\\qquad \\text{(E) }9 \\]\n4. Billy and Bob decide to build a rectangular field on the side of the river, i.e. so that they don't have to contribute any fencing to the side of the field against the river. Billy will only agree to contribute fencing to one side of the field that is perpendicular to the river, while Bob has agreed to contribute fencing to the other two sides. If Billy can contribute at most $50$ ft of fencing and Bob can contribute at most $70$ ft of fencing, what is the maximum possible area of their field?\n\\[\\text{(A) }870 ft^{2}\\qquad \\text{(B) }900 ft^{2}\\qquad \\text{(C) }1000 ft^{2}\\qquad \\text{(D) }1225 ft^{2}\\qquad \\text{(E) }1550 ft^{2}\\]\n5. What is $\\arg{\\left(\\frac{1}{\\frac{\\sqrt{3}}{2}+\\frac{i}{2}}+\\frac{1}{\\frac{1}{2}+\\frac{i\\sqrt{3}}{2}}\\right)}$?\n\\[\\text{(A) }0\\qquad \\text{(B) }\\frac{\\pi}{6}\\qquad \\text{(C) }\\frac{\\pi}{4}\\qquad \\text{(D) }\\frac{-\\pi}{4}\\qquad \\text{(E) }\\pi \\]\n6. What is $\\sum^{100}_{k=1}3k^{2}-3k+1$?\n\\[\\text{(A) }300301\\qquad \\text{(B) }99999\\qquad \\text{(C) }100000\\qquad \\text{(D) }999999\\qquad \\text{(E) }1000000 \\]\n7. Each user at Art of Problem Solving has some number of posts. If the number of people with at least 250 and fewer than 1000 posts is 332 more than the number of people with fewer than 250 posts, the number of people with at least 500 posts is 672, and the number of people with fewer than 250 posts or at least 1000 posts is 924, find the number of people with at least 250 but less than 500 posts.\n\\[\\text{(A) }80\\qquad \\text{(B) }508\\qquad \\text{(C) }584\\qquad \\text{(D) }782\\qquad \\text{(E) }1264 \\]\n8. A 6-letter license plate is formed by choosing three distinct letters from the alphabet as the first three letters of the license plate and then choosing three not necessarily distinct letters out of the first three to be the second three. How many license plates can be formed in this manner? For example, $ABCAAB$ works but $ABCABD$ does not.\n\\[\\text{(A) }46800\\qquad \\text{(B) }124200\\qquad \\text{(C) }140400\\qquad \\text{(D) }257400\\qquad \\text{(E) }421200 \\]\n9. $\\triangle ABC$ is a triangle such that $AB=5$, $AC=3$, and $BC=4$. Points $D$ and $E$ are on side $AB$ such that $AD=2$ and $BE=2$. Find the area of $\\triangle DEC$.\n\\[\\text{(A) }1\\qquad \\text{(B) }\\frac{6}{5}\\qquad \\text{(C) }\\sqrt{2}\\qquad \\text{(D) }2\\qquad \\text{(E) }\\frac{\\sqrt{5}}{12}\\]\n10. Let $a_{1}=2$ and $a_{n}=5a_{n-1}$ for $n \\ge 2$. If $\\ln{\\left(\\prod^{100}_{k=1}a_{k}\\right)}= a \\ln{2}+b \\ln{5}$, with $a$ and $b$ integers, what is the sum of the digits of $b-a$?\n\\[\\text{(A) }17 \\qquad \\text{(B) }18 \\qquad \\text{(C) }19 \\qquad \\text{(D) }20\\qquad \\text{(E) }21 \\]\n11. Let $G$ denote the triangle centroid of $\\triangle ABC$, and let $M_{A}$ denote the midpoint of $BC$. Given that $AB=6$, $BC=7$, and the area of $\\triangle G M_{A}B$ is $1$, what is the area of $\\triangle ABC$?\n\\[\\text{(A) }2\\sqrt{2}\\qquad \\text{(B) }\\sqrt{6}+\\sqrt{7}\\qquad \\text{(C) }5\\qquad \\text{(D) }6\\qquad \\text{(E) }7 \\]\n12. What is the graph of $(z+1)(\\overline{z}+1)=9$ in the complex plane?\n\\[\\text{(A)}\\text{point}\\qquad \\text{(B) }\\text{line}\\qquad \\text{(C) }\\text{circle}\\qquad \\text{(D) }\\text{two lines}\\qquad \\text{(E) }\\text{parabola}\\]\n13. If $f(x) = 2\\sin{\\left(\\frac{2 \\sqrt{2}x}{3}\\right)}+3\\cos{\\left(\\frac{3x}{2}\\right)}$, what is the number of values of p such that $0 that's something you can do just by putting in calculator. :rotfl: \r\n\r\nI'm quite hesistant to believe how some people in United States will get this one wrong using calculator. :D", + "Solution_12": "There is a mistake in #19.\r\n\r\nWe don't know what $a_{0}$ is. :maybe:", + "Solution_13": "Some people will get it wrong on purpose and some people will make bubbling errors. Some people may even misread the question. But yeah, it's better to at least make it a word problem. :)", + "Solution_14": "[quote=\"mathgeniuse^ln(x)\"]There is a mistake in #19.\n\nWe don't know what $a_{0}$ is. :maybe:[/quote]\r\n\r\nbut you know what $a_{1}$ is", + "Solution_15": "[quote=\"mathgeniuse^ln(x)\"]There is a mistake in #19.\n\nWe don't know what $a_{0}$ is. :maybe:[/quote]\r\n\r\nThere is no $a_{0}$, the sequence starts with $a_{1}$. Notice the sum starts with $k=1$.", + "Solution_16": "I'm soo in", + "Solution_17": "Hmm if I made no careless mistakes I got a 19*6+6*1.5=123. Does anyone else think there might be something wrong with #14?", + "Solution_18": "[quote=\"chess64\"]Hmm if I made no careless mistakes I got a 19*6+6*1.5=123. Does anyone else think there might be something wrong with #14?[/quote]\r\n\r\nWhat's wrong with 14? I got a perfectly reasonable answer.", + "Solution_19": "OK never mind then. I got 3 problems wrong though (105) :(", + "Solution_20": "[quote=\"Silverfalcon\"]calc rulz, please do not use problem like #1 again. Although some of you may be laughing, people DO miss #1 on real AMCs and those #1s are not 1+1 or 2+2 problems. Come on guys, be serious about making mock AMCs.[/quote]\r\n\r\nLast year, I scored highest in my state on the AMC 10 and I missed three problems - something like #1, #7, and #14. Everything else I answered correctly (so #25 correct, #1 wrong... :oops: )\r\n(Actually, now that I'm thinking it might have been #3 instead)\r\nI was rushing the easy ones and the word problem in that #1 (or 3) threw me off. It was something about ordering hamburgers and pizzas and stuff.\r\n\r\nThe point is, by not bothering to write word problems you are changing the nature of your test from the real thing.\r\n\r\nThe main point is, I wanted to back up Silverfalcon with an actual example. Besides that however, I really think your test is very good, I enjoyed taking it!", + "Solution_21": "[hide=\"comments\"]\nEDIT: By request of the writer. [/hide]", + "Solution_22": "I keep getting something that isn't a choice for 23, so I must be wrong :(", + "Solution_23": "[quote=\"bpms\"]I keep getting something that isn't a choice for 23, so I must be wrong :([/quote]\r\n\r\nYeah, I got one of the choices. For what that's worth.", + "Solution_24": "[quote=\"paladin8\"][quote=\"bpms\"]I keep getting something that isn't a choice for 23, so I must be wrong :([/quote]\n\nYeah, I got one of the choices. For what that's worth.[/quote]\r\nyou=genius\r\nme=idiot middle schooler\r\nI think that if you got one of the answer choices then I'm missing something...", + "Solution_25": "[quote=\"paladin8\"][hide=\"comments\"]\n[/hide][/quote]\r\n\r\nPlease do not post information like this while the contest is still going on.", + "Solution_26": "dammit, 120...thought i did a little better but i missed all the questions i guessed on (talk about luck) :(", + "Solution_27": "[quote=\"calc rulz\"]Please do not post information like this while the contest is still going on.[/quote]\r\n\r\nFair enough, I put it in the hide tag for a reason, but oh well. BTW, you might want to edit your post, too, or they can still read it there.", + "Solution_28": "[quote=\"Silverfalcon\"]chess64, you misunderstood my intention. I'm not angry or even raising my voice at anyone. It just came to me that in order for a good mock exam, it's much better to have a \"problem\" than mere calculations of 1+1 --> that's something you can do just by putting in calculator. :rotfl: \n\nI'm quite hesistant to believe how some people in United States will get this one wrong using calculator. :D[/quote]\r\n\r\ni guess i agree that the real test has to have some basic problems like that so everybody has something to do for 75 mins, but to be honest, the group that can only do those kinds of questions are unlikely to be taking mock tests on aops...so it does not really matter...", + "Solution_29": "Hmm Feiqi told me to bump it up if it's getting too far back in the AMC forum.", + "Solution_30": "I apologize if anyone saw what I posted a few minutes ago. That was completely unintentional... thank you altheman.", + "Solution_31": "Deadline is in 3 days!", + "Solution_32": "Answers due tomorrow!", + "Solution_33": "Is the contest over now? I'm curious to see how #24 is done ;)", + "Solution_34": "Nope, send your answers (and optionally solutions) in by today, I'll post answers tomorrow (I must say, I can't wait to post solutions, as a lot of people made many wrong assumptions).\r\n\r\nTime to grade people's submissions, I got kind of busy so anyone who submitted in the past few days hasn't gotten results back.", + "Solution_35": "is there somethign wrong with number 4?\r\nor am i crazy?", + "Solution_36": "[quote=\"deechou\"]is there somethign wrong with number 4?\nor am i crazy?[/quote]\r\n\r\nyou're crazy\r\n\r\njk", + "Solution_37": "Contest is over, answers and results to be posted later today.\r\n\r\n(ps: if you happen to send in answers like NOW, I might accept them)", + "Solution_38": "Contest is officially over." +} +{ + "Tag": [ + "trigonometry" + ], + "Problem": "Suppose that A,B,C are angles of triangle ABC and BC=a,CA=b,AB=c.\r\n1/Given $ 2\\widehat{A}\\plus{}3\\widehat{B}\\equal{}\\pi$. Prove that $ 4(a\\plus{}b)\\le 5c$\r\n2/Given $ tanA\\plus{}tanC\\equal{}2tanB$. Prove that $ cosA\\plus{}cosC\\le\\frac{3\\sqrt2}{4}$", + "Solution_1": "[hide]For the pro.2\ntanA+tanC=tanB\nSo sinAsinB=3cosAcosC(1)\nLet x=cosA,y=cosc(x,y>0)\n(1) is $ \\sqrt {1 \\minus{} x^2 } \\sqrt {1 \\minus{} y^2 } \\equal{} 3xy$\n $ \\Leftrightarrow (1 \\minus{} x^2 )(1 \\minus{} y^2 ) \\equal{} 9x^2 y^2$\n$ \\Leftrightarrow \\frac{9}{8} \\minus{} \\frac{1}{8}(8xy \\minus{} 1)^2 \\equal{} (x \\plus{} y)^2$ :) [/hide]\r\n :lol:", + "Solution_2": "My proof :lol: for no.1\r\nA+2B=C\r\n$ 4(\\sin A \\plus{} \\sin B) \\le 5\\sin (A \\plus{} 2B)$\r\n$ 4(\\sin A \\plus{} \\sin B) \\le 5(\\sin A(1 \\minus{} 2\\sin B^2 ) \\plus{} 2\\cos A\\sin B\\cos B)$\r\n$ 4\\sin B \\le 5\\sin \\frac{{180 \\minus{} 3B}}{2} \\plus{} 10\\cos \\frac{{180 \\minus{} B}}{2}\\sin B$\r\n$ 8\\cos \\frac{B}{2}\\sin \\frac{B}{2} \\le 4(1 \\minus{} \\sin \\frac{B}{2}^2 )\\cos \\frac{B}{2} \\minus{} 3\\cos \\frac{B}{2} \\plus{} 20\\sin \\frac{B}{2}^2 \\cos \\frac{B}{2}$\r\n$ 0 \\le \\cos \\frac{B}{2}(4\\sin \\frac{B}{2} \\minus{} 1)^2$\r\n :) Is that right?", + "Solution_3": "I think it's right!\r\nThank you for your nice solutions :)\r\nMy solution is very long and ugly!" +} +{ + "Tag": [ + "MATHCOUNTS" + ], + "Problem": "Please do not discuss specific problems or anything about the Chapter competition tests until March, as chapter competitions are held throughout February.\r\n\r\nDo not discuss State competitions until April.\r\n\r\nYou may discuss your placement, but do not discuss the competitions in any greater detail than that until the appropriate dates or the moderators will remove your comments.", + "Solution_1": "some people have not taken the school test yet, so it is best to wait a tid bit. you can post scores and stuffs though", + "Solution_2": "Our school's team's decided (finally)!", + "Solution_3": "so Chinaboy how did you do? You probably had automatic status after taking 3rd (I believe) in Oklahoma.", + "Solution_4": "he got a perfect score. and actually he did get 1st in the written round for Oklahoma last year", + "Solution_5": "a perfect score on a written test. that's like califronian people level I\"M AMAZED M.Y.", + "Solution_6": "erm, i don't remember what he got for the state test, but i know for sure he got perfect score on the school competition.", + "Solution_7": "i remeber (MC roomate) he got 2nd overall then went down to 3rd in oklahoma i think he got 40+ because i remebr he did better than me.\r\nwere you there syntax error?", + "Solution_8": "[quote=\"churchilljrhigh\"]i remeber (MC roomate) he got 2nd overall then went down to 3rd in oklahoma i think he got 40+ because i remebr he did better than me.\nwere you there syntax error?[/quote]\r\n\r\nwell, the person who got second and moved down to third was a girl, you may be referring to hao-fei who got 3rd and moved down to 4th. \r\n\r\nno, it was not me. i am proud of my 28 on the state test last year. but i have immensely improved...", + "Solution_9": "Our team was chosen yesterday. It's me, two of my first cousins, and one other 8th grade guy. We all live within 2 blocks of each other. There must be some inbreeding going on in Nebraska.\r\n\r\nWe only have taken the Sprint school test so far. I lucked out & didn't make any mistakes this year. The second highest on the Sprint was a 27, so I think we won't be terrible at Chapter.\r\n\r\nWe're taking the rest of the test as practice for Chapter in a couple weeks.\r\n\r\nPractices have actually relaxed a lot for some reason. Last Friday night we got together, had pizza and just did a bunch of Countdown rounds.\r\n\r\nAre your teams getting pretty serious yet?", + "Solution_10": "I can't comment on how serious the practices are going to be seeing as to how we just chose the team, but the team members are very serious. Our goal is top 5 team in nats this year. working for it at least", + "Solution_11": "chruchhill, i got a 40 last year, and i was bumped down from 1st to 2nd in countdown\r\n\r\nand neal, WELL SAID!\r\n\r\nlol.\r\n\r\nand perfect on school's not much of a big deal.\r\nIT'S SCHOOL!\r\n\r\nbut anyways, it was my first perfect...\r\n\r\nand then, amy's also on our team, and so is this person who got 8th in OK last year.\r\n\r\nand i know that hao fei roomed with josh and josh is not churchillhigh, lol.\r\n\r\nwell... enough \"news flashes\"... lol.", + "Solution_12": "I'm on the team! I'm happy, the team is me, two of my friends, and a guy who I used to like. We're all dedicated and smart, and we haven't gone to regionals yet, but it's my first time. Yay!!! I just found out two days ago.", + "Solution_13": "you were at states for cali... that was cool", + "Solution_14": "north cali!!", + "Solution_15": "I got a problem wrong because I cut 100 into four pieces of 50.", + "Solution_16": "dont complain yoo nationals person..." +} +{ + "Tag": [ + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "let $ X_n \\equal{} {t \\equal{} (t_1,t_2,t_3,...,t_n)| t_i \\equal{} 0 or t_i \\equal{} 1;}$\n\na) B is a subset of $ X_n ; |B| \\ge \\frac {2^{n \\plus{} 2}}{n}$ . Prove that there exists u,v,t in B such that d(u,v) = d(v,t) = d(t,v) denote d(u,v)= $ |u_1 \\minus{} t_1| \\plus{} |u_2 \\minus{} t_2| \\plus{} ... \\plus{} |u_n \\minus{} v_n|$ \n\nb) A is a subset of $ X_n$ such that u, v are distinct arbitrary in A; we have d(u,v)$ \\ge 3$ ; prove that $ |A| \\le \\frac {2^n}{n \\plus{} 1}$", + "Solution_1": "With typos, formatting, etc. fixed (something you should do youself in the future):\r\n\r\nLet $ X_n \\equal{} \\{t \\equal{} (t_1,t_2,t_3,...,t_n) \\mid t_i \\equal{} 0 \\text{ or } t_i \\equal{} 1\\} \\equal{} \\{0, 1\\}^n$. \r\n\r\na) $ B$ is a subset of $ X_n$ and $ |B| \\ge \\frac {2^{n \\plus{} 2}}{n}$. Prove that there exists $ u,v,t \\in B$ such that $ d(u,v) \\equal{} d(v,t) \\equal{} d(t,v)$, where we denote $ d(u,v) \\equal{} |u_1 \\minus{} v_1| \\plus{} |u_2 \\minus{} v_2| \\plus{} ... \\plus{} |u_n \\minus{} v_n|$. \r\n\r\nb) $ A$ is a subset of $ X_n$ and for every pair $ u, v$ of distinct elements of $ A$ we have $ d(u,v)\\ge 3$. Prove that $ |A| \\le \\frac {2^n}{n \\plus{} 1}$.", + "Solution_2": "[quote=\"JBL\"]Let $ X_n \\equal{} \\{t \\equal{} (t_1,t_2,t_3,...,t_n) \\mid t_i \\equal{} 0 \\text{ or } t_i \\equal{} 1\\} \\equal{} \\{0, 1\\}^n$. \n\na) $ B$ is a subset of $ X_n$ and $ |B| \\ge \\frac {2^{n \\plus{} 2}}{n}$. Prove that there exists $ u,v,t \\in B$ such that $ d(u,v) \\equal{} d(v,t) \\equal{} d(t,v)$, where we denote $ d(u,v) \\equal{} |u_1 \\minus{} v_1| \\plus{} |u_2 \\minus{} v_2| \\plus{} ... \\plus{} |u_n \\minus{} v_n|$.[/quote]\n\nI think $ \\left|B\\right| > \\frac {2^{n \\plus{} 1}}{n}$ would already suffice as a condition.\nWe call two elements $ x$ and $ y$ [i]adjacent[/i] if and only if $ d\\left(x,y\\right) \\equal{} 1$.\nFor any $ x\\in\\left\\{0,1\\right\\}^n$, let $ N\\left(x\\right)$ denote the set of all elements of $ \\left\\{0,1\\right\\}^n$ adjacent to $ x$.\nAssume that the assertion is false. Then, for any $ x\\in\\left\\{0,1\\right\\}^n$, there exist at most two elements of $ B$ which are adjacent to $ x$ (because if there would be three such elements $ u$, $ v$, $ t$, then they would satisfy $ d\\left(u,v\\right) \\equal{} d\\left(v,t\\right) \\equal{} d\\left(t,u\\right) \\equal{} 2$, and the assertion would be correct). In other words, for any $ x\\in\\left\\{0,1\\right\\}^n$, there exist at most two elements $ y$ of $ B$ such that $ x\\in\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}$.\nThus, $ \\sum_{y\\in B}\\left|\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}\\right|\\leq 2\\left|\\left\\{0,1\\right\\}^n\\right|$. But\n\n$ \\sum_{y\\in B}\\underbrace{\\left|\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}\\right|}_{ \\equal{} n} \\equal{} \\sum_{y\\in B}n \\equal{} \\left|B\\right|\\cdot n$ and $ \\left|\\left\\{0,1\\right\\}^n\\right| \\equal{} 2^n$,\n\nso this becomes $ \\left|B\\right|\\cdot n\\leq 2\\cdot 2^n$, contradicting $ \\left|B\\right| > \\frac {2^{n \\plus{} 1}}{n}$.\n\n[quote=\"JBL\"]b) $ A$ is a subset of $ X_n$ and for every pair $ u, v$ of distinct elements of $ A$ we have $ d(u,v)\\ge 3$. Prove that $ |A| \\le \\frac {2^n}{n \\plus{} 1}$.[/quote]\r\n\r\nTry [url=http://en.wikipedia.org/wiki/Hamming_bound]Hamming bound on Wikipedia[/url].\r\n\r\n darij", + "Solution_3": "Part (a) of this problem is a reworded version of (a) in [url=http://www.mathlinks.ro/viewtopic.php?t=143942]Romanian TST 2 2007, Problem 4[/url]", + "Solution_4": "[quote=\"darij grinberg\"]\n\nI think $ \\left|B\\right| > \\frac {2^{n \\plus{} 1}}{n}$ would already suffice as a condition.\nWe call two elements $ x$ and $ y$ [i]adjacent[/i] if and only if $ d\\left(x,y\\right) \\equal{} 1$.\nFor any $ x\\in\\left\\{0,1\\right\\}^n$, let $ N\\left(x\\right)$ denote the set of all elements of $ \\left\\{0,1\\right\\}^n$ adjacent to $ x$.\nAssume that the assertion is false. Then, for any $ x\\in\\left\\{0,1\\right\\}^n$, there exist at most two elements of $ B$ which are adjacent to $ x$ (because if there would be three such elements $ u$, $ v$, $ t$, then they would satisfy $ d\\left(u,v\\right) \\equal{} d\\left(v,t\\right) \\equal{} d\\left(t,u\\right) \\equal{} 2$, and the assertion would be correct). In other words, for any $ x\\in\\left\\{0,1\\right\\}^n$, there exist at most two elements $ y$ of $ B$ such that $ x\\in\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}$.\nThus, $ \\sum_{y\\in B}\\left|\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}\\right|\\leq 2\\left|\\left\\{0,1\\right\\}^n\\right|$. But\n\n$ \\sum_{y\\in B}\\underbrace{\\left|\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}\\right|}_{ \\equal{} n} \\equal{} \\sum_{y\\in B}n \\equal{} \\left|B\\right|\\cdot n$ and $ \\left|\\left\\{0,1\\right\\}^n\\right| \\equal{} 2^n$,\n\nso this becomes $ \\left|B\\right|\\cdot n\\leq 2\\cdot 2^n$, contradicting $ \\left|B\\right| > \\frac {2^{n \\plus{} 1}}{n}$.\n[/quote]\r\nDear darij grinberg!\r\nYour solution is very good. But I think there are a little thing in it. I don't think $ \\left|\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}\\right|$ equals to $ n$. I've tried in the case $ y \\equal{} (1,0,1)$, and $ t\\in \\{(1,0,0),(1,1,1),(0,0,1),(0,1,0)\\}$ then $ d(y,t) \\equal{} 1$! But your solution is still ok. It is clearly that $ \\left|\\left\\{t\\in\\left\\{0,1\\right\\}^n\\mid d\\left(y,t\\right) \\equal{} 1\\right\\}\\right|\\ge n.$", + "Solution_5": "Thanks, but no: $ d\\left(\\left(1,0,1\\right),\\left(0,1,0\\right)\\right)$ is $ 3$, not $ 1$ (we are not working modulo $ 2$).\r\n\r\n darij", + "Solution_6": "[quote=\"darij grinberg\"]Thanks, but no: $ d\\left(\\left(1,0,1\\right),\\left(0,1,0\\right)\\right)$ is $ 3$, not $ 1$ (we are not working modulo $ 2$).\n\n darij[/quote]\r\nOh, I am sorry. Thankyou very much. I've mistake that $ d(u,v)\\equal{}|(u_1\\minus{}v_1)\\plus{} (u_2\\minus{}v_2)\\plus{}...\\plus{}(u_n\\minus{}v_n)|$! I will be careful more. Thanks again" +} +{ + "Tag": [], + "Problem": "Hello Guys,\r\n\r\nI am a swedish high school student, interested in studiying at a major american college. Most of the reknown colleges emphazise taking the SATs (1 and 2) before January. \r\n\r\nI have to take the SAT Subject tests in December since I am participating in the Baltic Way Mathematical Competition during the November SAT date. \r\n\r\nThe thing is that I am going to the USA for a wedding in October (during the SAT october date), and I am going to attend the ceremony in the evening of the 9th of October (the exact day of the SAT). Anyhow, my plan is to take the SAT1 before the wedding ceremony. Is this possible (does the test start, let's say, before 14:00) ?\r\n\r\nBest Regards,", + "Solution_1": "The test starts at 8:00, and should end at about 12:00.", + "Solution_2": "Just so you know, you have to register in advance. And soon. Go to collegeboard.com to do that. You cannot just show up to the test and take it.", + "Solution_3": "Okey Magnara, I already knew that, but thanks for pointing it out anyway. I will sign up during or after the weekend.\r\n\r\nRegards", + "Solution_4": "[quote=\"arccosinus\"]I am a swedish high school student, interested in studiying at a major american college.[/quote]\r\n\r\nCheck the [url=http://www.collegeboard.com/?student]College Board Web site[/url] for rules about test-takers from different countries. It is worthwhile to check, for instance, whether or not you can take the test at all outside the country you live in. You will definitely want to have your passport along as photo identification if you take the test in the United States. \r\n\r\nAnd, yes, as another participant suggested, check registration deadlines right away. \r\n\r\nGood luck!" +} +{ + "Tag": [ + "inequalities solved", + "inequalities" + ], + "Problem": "Let $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ be positive real numbers such that $ a \\plus{} x \\equal{} b \\plus{} y \\equal{} c \\plus{} z \\equal{} 1$.\r\n\r\nProve that\r\n\r\n $ \\left(abc \\plus{} xyz\\right)\\left(\\frac {1}{ay} \\plus{} \\frac {1}{bz} \\plus{} \\frac {1}{cx}\\right)\\geq 3$.", + "Solution_1": "Extremely beautiful:\r\n Observe that abc+(1-a)(1-b)(1-c)=1-a-b-c+ab+bc+ca=(1-b)(1-c)+ca+ab-a. Using this type of identities we obtain that \r\n (abv+(1-a)(1-b)(1-c))*sum 1/a(1-b)=a/(1-c)+b/(1-a)+c/(1-b)+(1-c)/a+(1-a)/b+(1-b)/c. Apply now AM-GM for these 6 numbers." +} +{ + "Tag": [ + "number theory proposed", + "number theory" + ], + "Problem": "Let be p>5 a prime number. Prove that exists a and b ( with a^2 < p and b^2 < p ) such that p-a^2 is a divisor of p-b^2", + "Solution_1": "$a=b<\\sqrt p$.", + "Solution_2": "i'm sory! I forgot to say that a can't be equal with b.", + "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?p=495423#p495423\r\nBalkan MO :wink:" +} +{ + "Tag": [ + "inequalities", + "function", + "rearrangement inequality", + "inequalities solved", + "3-variable inequality", + "cyclic inequality" + ], + "Problem": "Prove that for all positive real numbers $a,b,c$ the following inequality takes place \r\n\\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{27}{2(a+b+c)^2} . \\] \r\n[i]Laurentiu Panaitopol, Romania[/i]", + "Solution_1": "The first one is very easy:\r\n1/c(a+b)+1/b(a+c)+1/a(b+c) \\geq 27/2(a+b+c) 2 \r\n9/2(ab+bc+ca) \\geq 27/2(a+b+c) 2 \r\n1/(ab+bc+ca) \\geq 3/(a+b+c) 2 \r\n(a+b+c) 2 \\geq 3(ab+bc+ca) which is obviously true\r\n\r\n\r\n\r\nCheers :)", + "Solution_2": "Rewrite the inequality like:\n\\[\\left[\\sum_{cyc} (a+b)\\right]\\left[\\sum_{cyc} \\frac1{b(a+b)}\\right]\\left[\\sum_{cyc} a\\right] \\ge 27\\]\nand apply AM -GM to the three sums and the inequality is solved!!", + "Solution_3": "First inequality is :\r\n\r\n1/b(a+b) + 1/c(b+c) + 1/a(c+a) \\geq .......\r\n\r\nand not\r\n\r\n1/a(b+c) + 1/b(a+c) +1/c(a+b) \\geq ........\r\n\r\nI think it is a different inequality.", + "Solution_4": "hmm.. but my method still works, doesn't it?", + "Solution_5": "Lagrangia, you are right and your proof is very neat but I referred to\r\nMaverik's proof.\r\nMy proof is:\r\nchanging a with b the inequality remains valid and is transformed in\r\n\r\n(1/a(b+a)) +(1/c(a+c))+(1/b(b+c)) \\geq ......\r\n\r\nsumming the two inequalities we obtain\r\n\r\n (a+b+c)/abc \\geq 27/(a+b+c)^2\r\n\r\nwhic can be rewritten as\r\n\r\n (a+b+c)^3 \\geq 27abc\r\n\r\ntrue by AM-GM inequality.", + "Solution_6": "Yeah sorry about that!\r\nBut using rearangement inequality we get:\r\n1/b(a+b)+1/c(b+c)+1/a(a+c) \\geq 1/c(a+b)+1/a(b+c)+1/b(a+c)\r\nso then we use \\sum 1/c(a+b) \\geq 9/2(ab+bc+ca) ........ and so on.....", + "Solution_7": "Another solution is :\r\nFrom AM-GM ineq we have \r\n$\\frac{2}{b(a+b)}+\\frac{2}{c(b+c)}+\\frac{2}{a(c+a)}\\geq\\frac{6}{XY}$\r\n\r\nwhere $X=\\sqrt[3]{abc}$ and $Y=\\sqrt[3]{(a+b)(b+c)(c+a)}$ \r\n\r\nBy AM-GM again we have that $X\\leq\\frac{a+b+c}{3}$ and $Y\\leq\\frac{2a+2b+2c}{3}$\r\n\r\nThe result follows", + "Solution_8": "Yes, AM-Gm twice, Cauchy, rearangement, or direct calculations work on this inequality.\r\nAnyway, this inequality has been proposed by Greece, not by L. Panaitopol . ;)", + "Solution_9": "Yes exactly, but I was afraid to tell this. But now I am sure. The proposer is \r\n\r\nSotiris Louridas. \r\n\r\nAdditionally the best solution is the one which use Caushy in Engel form. I anyone want it i will post it", + "Solution_10": "The Cauchy solution is well-known. Actually, I have found a nice solution using rearangement inequality and then, I reduced it to Nesbitt. ;)", + "Solution_11": "Could you post it ? thanks. But with details. With no use of trivial etc", + "Solution_12": "Well, by rearangement we have \\[ \\frac{1}{b(a+b)}+\\frac{1}{c(b+c)}+\\frac{1}{a(c+a)}\\geq \\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}. \\]\n\nActually we will prove this stronger inequality: \\[ \\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}\\geq\\frac{9}{2(ab+bc+ca)}. \\]\n\nWell, this las inequality is equivalent with \\[ \\frac{ab+bc+ca}{a(b+c)}+\\frac{ab+bc+ca}{b(c+a)}+\\frac{ab+bc+ca}{c(a+b)}\\geq\\frac{9}{2}. \\]\n This last inequality is \\[ \\frac{bc}{a(b+c)}+\\frac{ca}{b(c+a)}+\\frac{ab}{c(a+b)}\\geq\\frac{3}{2}. \\]\n\n Well, by making the substitutions $bc=m, ca=n$ and $ca=p$ the inequality finally reduces to\n \n \\[ \\frac{m}{n+p}+\\frac{n}{p+m}+\\frac{p}{m+n}\\geq\\frac{3}{2}. \\]\n\n From here you can apply Nesbitt to finish.", + "Solution_13": "I hope this solution is in detail. ;)", + "Solution_14": "[quote=\"cezar lupu\"]I hope this solution is in detail. ;)[/quote]\r\n\r\nMaybe the only of yours :rotfl:", + "Solution_15": "[quote=starcraft][quote=\"cezar lupu\"]Well, by rearangement we have \\[ \\frac{1}{b(a+b)}+\\frac{1}{c(b+c)}+\\frac{1}{a(c+a)}\\geq \\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}. \\]\n\nActually we will prove this stronger inequality: \\[ \\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}\\geq\\frac{9}{2(ab+bc+ca)}. \\]\n\nWell, this las inequality is equivalent with \\[ \\frac{ab+bc+ca}{a(b+c)}+\\frac{ab+bc+ca}{b(c+a)}+\\frac{ab+bc+ca}{c(a+b)}\\geq\\frac{9}{2}. \\]\n This last inequality is \\[ \\frac{bc}{a(b+c)}+\\frac{ca}{b(c+a)}+\\frac{ab}{c(a+b)}\\geq\\frac{3}{2}. \\]\n\n Well, by making the substitutions $bc=m, ca=n$ and $ca=p$ the inequality finally reduces to\n \n \\[ \\frac{m}{n+p}+\\frac{n}{p+m}+\\frac{p}{m+n}\\geq\\frac{3}{2}. \\]\n\n From here you know how to proceed. ;)[/quote]\n\nYou are perfect.I admire you very much.[/quote]\n\nNice", + "Solution_16": "[quote=manlio]Prove that for all positive real numbers $a,b,c$ the following inequality takes place \n\\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{27}{2(a+b+c)^2} . \\] \n[i]Laurentiu Panaitopol, Romania[/i][/quote]\n[Hide=one-liner]\nBy AM-GM,\n$$\\frac{2}{b(a+b)}+\\frac{27(a+b)}{2(a+b+c)^3}+\\frac{27b}{(a+b+c)^3}\\geq\\frac{27}{(a+b+c)^2}$$\nAdding similar inequalities we get the required result", + "Solution_17": "By Rearrangement we have \\[ \\frac{1}{b(a+b)}+\\frac{1}{c(b+c)}+\\frac{1}{a(c+a)}\\geq \\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)} \\] also by Cauchy-Schwarz and AM-GM we have \\[ 2(a+b+c)^2(\\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}) \\geq 6(ab+ac+bc)(\\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)})=3[a(b+c)+b(c+a)+c(a+b)](\\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)}) \\geq 3(1+1+1)^2=27 \\] then $$\\frac{1}{a(b+c)}+\\frac{1}{b(c+a)}+\\frac{1}{c(a+b)} \\geq \\frac{27}{2(a+b+c)^2}$$", + "Solution_18": "[quote=sqing][quote=manlio]Prove that for all positive real numbers $a,b,c$ the following inequality takes place \\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{27}{2(a+b+c)^2} . \\] \n[i]Laurentiu Panaitopol, Romania[/i][/quote]\n we will prove this stronger inequality\n$\\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq\\frac{9}{2(ab+bc+ca)}$\n$\\iff\\frac{2(ab+bc+ca)}{b(a+b)}+\\frac{2(ab+bc+ca)}{c(b+c)}+\\frac{2(ab+bc+ca)}{a(c+a)}\\geq 9.$\nBy AM-GM,\n$\\sum\\frac{2(ab+bc+ca)}{b(a+b)}= \\sum\\frac{a+c}{a+b}+\\sum\\frac{c}{b}+\\sum\\frac{a(b+c)}{b(a+b)}\\geq3+3+3=9.$[/quote]\n[url=https://artofproblemsolving.com/community/c6h2334983p18769745]2009 Belarus:[/url]\nProve that any positive real numbers a,b,c satisfy the inequlaity$$\\frac{1}{(a+b)b}+\\frac{1}{(b+c)c}+\\frac{1}{(c+a)a}\\ge \\frac{9}{2(ab+bc+ca)}$$", + "Solution_19": "Here is my solution i know there above correct solution also but i m just sending this for just store it and someone might understand this if something wrong in this please let me know :D \n[b][color=#00f]My solution:[/color][/b]\\\\\nWe can see that here is very standard Problem So Lets do it\\\\\nLets Apply AM-GM here,\n\\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{3}{\\sqrt[3]{abc((a+b)(b+c)(c+a))}}\\]\nSo to make it Simple\\\\\nLet $X=\\sqrt[3]{abc}$ and $Y=\\sqrt[3]{(a+b)(b+c)(c+a)}$ So now its looks like \\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{3}{XY}\\] now its looks nice then Apply some AM-GM Lets see what we find \\[3\\sqrt[3]{abc}=3X\\leq a+b+c\\implies X\\leq \\frac{a+b+c}{3}\\] Similarly \\[3\\sqrt[3]{(a+b)(b+c)(c+a)}=3Y\\leq a+b+b+c+c+a=2(a+b+c)\\implies Y\\leq \\frac{2(a+b+c)}{3}\\]so we can clearly see that \\[XY\\leq \\frac{2(a+b+c)^2}{9}\\]\\[\\implies\\frac{1}{XY}\\geq \\frac{9}{2(a+b+c)^2}\\]\nNow we have to Just plug it Up and we are done Demon Killed.\n\\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{3}{\\sqrt[3]{abc((a+b)(b+c)(c+a))}}\\geq \\frac{3}{XY}\\geq \\frac{3.9}{2(a+b+c)^2}\\]\n\\[\\implies \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)}\\geq \\frac{27}{2(a+b+c)^2}\\]\nAnd we are done demon killed.Mission Accomplished. :-D ", + "Solution_20": "Nice one@above", + "Solution_21": "[hide=Storage]We use the Rearrangement Inequality along with the Cauchy Schwarz Inequality :) \nSince the inequality is cyclic we need to consider 2 cases namely when\n$$ 1) a \\ge b \\ge c$$\nNote that from the above condition we get that $\\frac{1}{a} \\le \\frac{1}{b} \\le \\frac{1}{c}$ and $\\frac{1}{a+b} \\le \\frac{1}{a+c} \\le \\frac{1}{b+c}$ \nTherefore by the Rearrangement Inequality and the Cauchy Schwarz Inequality (Titu's lemma) we can conclude that $$\\sum_{cyc} \\frac{1}{b(a+b)} \\ge \\sum_{cyc} \\frac {1}{c(a+b)} \\ge \\frac {(1+1+1)^2}{2(ab+bc+ca)} = \\frac{9}{2(ab+bc+ca)} \\ge \\frac{9}{2\\frac{(a+b+c)^2}{3}} = \\frac{27}{2(a+b+c)^2}$$\n$$2) a \\ge c \\ge b$$ \nWe can similarly deduce the same result since this case is very similar to the above one and so we are done! :D \n[/hide]\n[hide=Remarks]I have not used the Rearrangement Inequality very often and I am a noob so please tell if there is any error in my solution :maybe: [/hide]", + "Solution_22": "[quote=sqing][hide][quote=sqing][quote=manlio]Prove that for all positive real numbers $a,b,c$ the following inequality takes place \\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{27}{2(a+b+c)^2} . \\] \n[i]Laurentiu Panaitopol, Romania[/i][/quote]\n we will prove this stronger inequality\n$\\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq\\frac{9}{2(ab+bc+ca)}$\n$\\iff\\frac{2(ab+bc+ca)}{b(a+b)}+\\frac{2(ab+bc+ca)}{c(b+c)}+\\frac{2(ab+bc+ca)}{a(c+a)}\\geq 9.$\nBy AM-GM,\n$\\sum\\frac{2(ab+bc+ca)}{b(a+b)}= \\sum\\frac{a+c}{a+b}+\\sum\\frac{c}{b}+\\sum\\frac{a(b+c)}{b(a+b)}\\geq3+3+3=9.$[/quote]\n[url=https://artofproblemsolving.com/community/c6h2334983p18769745]2009 Belarus:[/url]\nProve that any positive real numbers a,b,c satisfy the inequlaity$$\\frac{1}{(a+b)b}+\\frac{1}{(b+c)c}+\\frac{1}{(c+a)a}\\ge \\frac{9}{2(ab+bc+ca)}$$[/quote]\n\nThat's a easy one. Apply Titu's lemma and done.", + "Solution_23": "@above Titu doesn't work here directly.\n\n[quote=sqing][url=https://artofproblemsolving.com/community/c6h2334983p18769745]2009 Belarus:[/url]\nProve that any positive real numbers a,b,c satisfy the inequlaity$$\\frac{1}{(a+b)b}+\\frac{1}{(b+c)c}+\\frac{1}{(c+a)a}\\ge \\frac{9}{2(ab+bc+ca)}$$ [/quote]\n\nNote that it's equivalent to $$2\\sum_{\\textsf{cyc}}{a^{4}b^{2}} + 2\\sum_{\\textsf{cyc}}{a^{4}bc}+2\\sum_{\\textsf{cyc}}{a^{3}b^{3}}\\ge a^{3}b^{2}c+b^{3}c^{2}a+c^{3}a^{2}b+3\\sum_{\\textsf{cyc}}{a^{3}bc^{2}}+6a^{2}b^{2}c^{2}$$ But \\begin{align*} 2\\sum_{\\textsf{cyc}}{a^{4}b^{2}}\\ge 2\\sum_{\\textsf{cyc}}{a^{3}bc^{2}} \\hspace{2 mm} \\textsf{and} \\hspace{2 mm} (4, 1, 1)\\ge (2, 2, 2) \\end{align*} So it remains to show $$(3, 3, 0)\\ge (3, 2, 1)\\hspace{2 mm} \\textsf{which is true by Muirhead.}$$", + "Solution_24": "Many unnecessarily lengthy solutions and how is \u201cHolder\u201d never mentioned in the thread - posting for strorage.\n\n$$(\\frac{1}{b(a+b)}+\\frac{1}{c(b+c)}+\\frac{1}{a(a+c)})(b+c+a)((a+b)+(b+c)+(a+c)) \\geq 27$$\nby Holder\u2019s Inequality and the conclusion follows readily.\n", + "Solution_25": "Instant Holder...\n\\begin{align*}\n \\left(\\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)}\\right) \\cdot (a+b+c)\\cdot (a+b+b+c+c+a)\\geq 3^3=27.\n\\end{align*}\n", + "Solution_26": "[quote=manlio]Prove that for all positive real numbers $a,b,c$ the following inequality takes place \n\\[ \\frac{1}{b(a+b)}+ \\frac{1}{c(b+c)}+ \\frac{1}{a(c+a)} \\geq \\frac{27}{2(a+b+c)^2} . \\] \n[i][color=#f00]Laurentiu Panaitopol, Romania[/color][/i][/quote]\n\nplease correct the proposer (as well as the title) to: \n\n[i]Sotiris Louridas, Greece[/i] \n\n(see posts #09,10)", + "Solution_27": "Let $a, b, c$ be positive real numbers. Prove that \n$$\\dfrac{1}{b(a + 2b)} + \\frac{1}{c(b +2 c)}+ \\frac{1}{a(c + 2a)} \\ge \\frac{9}{(a + b + c)^2}$$\n$$\\dfrac{1}{b(a +3b)} + \\frac{1}{c(b +3c)}+ \\frac{1}{a(c +3a)} \\ge \\frac{27}{4(a + b + c)^2}$$ ", + "Solution_28": "Let $a, b, c$ be positive real numbers. Prove that \n$$\\dfrac{1}{b(a +kb)} + \\frac{1}{c(b +kc)}+ \\frac{1}{a(c +ka)}\\ge \\frac{27}{(k+1)(a + b + c)^2}$$\nWhere $k>0.$", + "Solution_29": "[b]Generalization 1[/b]\nLet $a_{1}, a_{2},\\cdots,a_{2k+1}$ be positive reels. Then prove that\n\n$$\\sum_{n=1}^{2k+1}{\\dfrac{1}{a_{n}\\left(a_{n-2k+1}+a_{n-2k+2}+\\cdots+a_{n}\\right)}}\\geq \\dfrac{(2k+1)^3}{2k\\left(\\sum_{cyc}{a_{k}}\\right)}$$" +} +{ + "Tag": [ + "function" + ], + "Problem": "Find all functions $ f: \\mathbb R \\minus{} \\{0\\}\\rightarrow\\mathbb R \\minus{} \\{0\\}$ such that $ f(1) \\equal{} 1$ and $ f(yf(x) \\plus{} \\frac {x}{y}) \\equal{} xyf(x^2 \\plus{} y^2)$ for all $ x,y\\in\\mathbb R \\minus{} \\{0\\}$.", + "Solution_1": "[hide]One solution is $ f(x) \\equal{} \\frac1x$. But, is the following reasoning legitimate?\n\nLet's consider x = 1 in the functional equation. Then we get\n\n$ f[y \\cdot f(1) \\plus{} \\frac1y] \\equal{} y \\cdot f(1\\plus{}y^2)$\n\n$ \\equal{}> f\\left(y \\plus{} \\frac1y\\right) \\equal{} y \\cdot f(1\\plus{}y^2)$\n\n$ \\equal{}> f\\left(\\frac{1\\plus{}y^2}{y}\\right) \\equal{} y \\cdot f(1\\plus{}y^2)$\n\nConsider now $ 1 \\plus{} y^2 \\equal{} u$. Then, for u = 2y we get\n\n$ f(2) \\equal{} \\frac{u}{2} \\cdot f(u) \\equal{}> f(u) \\equal{} \\frac{2 \\cdot f(2)}{u}$, which implies $ f(x) \\equal{} \\frac1x$.[/hide]", + "Solution_2": "[quote=\"Carcul\"]One solution is $ f(x) \\equal{} \\frac1x$. But, is the following reasoning legitimate?\n\nLet's consider x = 1 in the functional equation. Then we get\n\n$ f[y \\cdot f(1) \\plus{} \\frac1y] \\equal{} y \\cdot f(1 \\plus{} y^2)$\n\n$ \\equal{} > f\\left(y \\plus{} \\frac1y\\right) \\equal{} y \\cdot f(1 \\plus{} y^2)$\n\n$ \\equal{} > f\\left(\\frac {1 \\plus{} y^2}{y}\\right) \\equal{} y \\cdot f(1 \\plus{} y^2)$\n\nConsider now $ 1 \\plus{} y^2 \\equal{} u$. Then, for u = 2y we get\n\n$ >>> f(2) \\equal{} \\frac {u}{2} \\cdot f(u) \\equal{} > f(u) \\equal{} \\frac {2 \\cdot f(2)}{u}$, which implies $ f(x) \\equal{} \\frac1x <<<$.[/quote]\r\nhow can you know that $ f(2)\\equal{}\\frac{1}{2}$ ?? and i think it must be\r\nConsider now $ 1 \\plus{} y^2 \\equal{} u$. Then, for $ u \\equal{} 2y$ we get\r\n$ f(\\frac{4\\plus{}u^2}{2u})\\equal{}\\frac{u}{2}f(\\frac{4\\plus{}u^2}{4})$\r\nanyway this problem is very hard :( I really want a solution for it ..." +} +{ + "Tag": [ + "function", + "algebra proposed", + "algebra" + ], + "Problem": "$n$ is a positive integer, $K$ the set of polynoms of real variables $x_1,x_2,...,x_{n+1}$ and $y_1,y_2,...,y_{n+1}$, function $f:K\\rightarrow K$ satisfies\r\n\\[f(p+q)=f(p)+f(q),\\quad f(pq)=f(p)q+pf(q),\\quad (\\forall)p,q\\in K.\\]\r\nIf $f(x_i)=(n-1)x_i+y_i,\\quad f(y_i)=2ny_i$ for all $i=1,2,...,n+1$ and\r\n\\[\\prod_{i=1}^{n+1}(tx_i+y_i)=\\sum_{i=0}^{n+1}p_it^{n+1-i}\\]\r\nfor any real $t$, prove, that for all $k=1,...,n+1$\r\n\\[f(p_{k-1})=kp_k+(n+1)(n+k-2)p_{k-1}\\]", + "Solution_1": "Sasha wrote $n$ instead $n+1$. So I think affirmation that we need to demonstrate need to be changed.\r\n\r\nThe original problem was:\r\n$n$ is a positive integer, $K$ the set of polynoms of real variables $x_1,x_2,...,x_{n+1}$ and $y_1,y_2,...,y_{n+1}$, function $f:K\\rightarrow K$ satisfies\r\n\\[f(p+q)=f(p)+f(q),\\quad f(pq)=f(p)q+pf(q),\\quad (\\forall)p,q\\in K.\\]\r\nIf $f(x_i)=(n-1)x_i+y_i,\\quad f(y_i)=2ny_i$ for all $i=1,2,...,n+1$ and\r\n\\[\\prod_{i=1}^{n+1}(tx_i+y_i)=\\sum_{i=0}^{n+1}p_it^{n+1-i}\\]\r\nfor any real $t$, prove, that for all $k=1,...,n+1$\r\n\r\n$f(p_{k-1})=kp_k+(n+1)(n+k-2)p_{k-1}$" +} +{ + "Tag": [ + "algebra", + "polynomial", + "calculus", + "integration", + "Rational Root Theorem" + ], + "Problem": "Show that the polynomial [tex]x^3+(5m+2)x+(5n+1)[/tex] does not have rational root for any integers [tex]m[/tex] and [tex]n[/tex].", + "Solution_1": "[hide]The only rational roots are :pm: p/q = :pm: (5n+1)\n\n\n\nUse synthetic division to divide those roots into the polynomial, and neither have remainder=0.[/hide]", + "Solution_2": "The rational roots are integers which are divisors of (5n+1).", + "Solution_3": "why must they be integers?", + "Solution_4": "The Rational Root Theorem. [quote=\"probability1.01\"]why must they be integers?[/quote]", + "Solution_5": "Well I think m and n have to be rational numbers. If they weren't then p and q wouldent be integers, and p/q would not be an integer, and thus they would not be RATIONAL, which they must be for the RATIONAL root theorem.", + "Solution_6": "[quote=\"JS1527\"]Well I think m and n have to be rational numbers...[/quote]\r\n\r\nThey are integers by assumption.", + "Solution_7": "Right. But if you wanted to apply the rational root theorem, w/o that assumption, they would have to be rational numbers.", + "Solution_8": "Yes, that is correct. However, did you see the error in your method that someone else pointed out?", + "Solution_9": "Refering to the very first reply. Since 5n+1 could have many factors that need to be tested, I don't see how you can use synthetic division to test them since they are unknown. \n\n\n\nHint:\n\n[hide]I am not convinced that the Rational Root theorem is useful here. The terms 5m+2 and 5n+1 suggest that mod 5 arithmetic might be useful somehow. [/hide]", + "Solution_10": "a hint: [hide]if a polynomial has a root in the integers, then it should have a root mod m for any m.[/hide]", + "Solution_11": "I am VERY bad when it comes to proving stuff, but here it goes:\n\n\n\n[hide]\n\nIf x^3 + 5mx + 2x + 5n + 1 = 0, then x^3 + 5mx + 2x +5n + 1 :equiv: x^3 + 2x + 1 :equiv: 0 (mod 5). From the Rational Root Theorem, the only possible rational solutions to x^3 + 2x + 1 are :pm: 1. Since those do not work, there are no rational solutions to x^3 + 5mx + 2x + 5n + 1 = 0.\n\n[/hide]\n\n\n\nI just know I made a mistake somewhere...", + "Solution_12": "A good try, but you've got one problem -- imagine that we applied the polynomial x^3 + 2x + 1, when, say, x = 3, and got, say, 75. (Yeah, we don't, but let's play along.) This would be congruent to 0 mod 5. Thus, the rational roots theorem isn't what is needed here. A good first step, however.", + "Solution_13": "I always have that problem and can't find a way to get around it.", + "Solution_14": "Well, here's the thing: the problem with the original problem is that you would have needed to plug in some unknown number of numbers into x to see if they were zeroes. But we're in mod 5 now -- how many different numbers do you have to plug in to check?", + "Solution_15": "Five - the numbers 0 - 4 because the numbers above 4 are congruent to 0, 1, 2, 3, or 4 mod 5.", + "Solution_16": "Excellent. And since by the rational root theorem, all rational roots are integral, you can simply check those 5 values yourself.", + "Solution_17": "I checked and none are zeroes. Since there are no zeroes for that equation, there are no solutions for the first one. QED", + "Solution_18": "Joel and nr1337:\r\n\r\nI am convinced by the second proof.\r\n\r\nHowever Joel, since your 'counter-example' was not one, I am still not convinced that the nr1377's first proof was incorrect. Ie. by checking the same five cases, it is proved that x^3+2x+1 never equals 0mod5 for integral x. So a counter-example would have to use a different polynomial and/or a different base - can anyone find one? Or is the Rational Root Theorem extendable to modn (if leading coefficient is not 1, things are messy; for now let's let it equal 1 and the Rational Root Theorem temporarily is the \"Integral Root Theorem\").\r\n\r\nNice extension!", + "Solution_19": "I am not sure this is what you are asking, but consider the equation x2 + 1 :equiv: 0 mod 5. If we tried to use something like the Rational Root Theorem, we would consider only x = :pm: 1, which are not solutions. But x = 2 is a solution.", + "Solution_20": "yes, but mod 5 the residues 1, 2, 3, 4 all divide 1 so it seems to me that you would still need to try each.", + "Solution_21": "I guess I don't understand what Yooper's question is.", + "Solution_22": "I think also that the rational root theorem can only get weirder mod m for m not a prime, because in this case if the degree of the polynomial is n, it can have more than n roots.", + "Solution_23": "Yes, Ravi, that's all I was asking. You found a very simple and convincing counter-example. Without thinking too much, I was just wondering wheher this train of thought would lead anywhere.\r\n\r\nThanks" +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "Prove that for all distint positive real numbers $a,b,c$ \\[\\frac{(a-b)^{2}}{(a-c)^{2}}+\\frac{(b-c)^{2}}{(b-a)^{2}}+\\frac{(c-a)^{2}}{(c-b)^{2}}\\ge \\frac{a+b}{a+c}+\\frac{b+c}{b+a}+\\frac{c+a}{c+b}.\\]", + "Solution_1": "the only way i see to solve this involves uglification, which i don't want to do. but i think it is solvable after multiplying through by (a-c)^2(b-c)^2(b-a)^2", + "Solution_2": "Please give me a solution. It's better.", + "Solution_3": "[b]Not[\\b] [b]complete[\\b] solution.\r\nFirst:I will prove a lemma.\r\n$Lemma$,if $a,b,c>0$,and $00$).\r\n(1) Solve the inequality: $f(x) \\leq 1$.\r\n(2) What value does $a$ take to make $f(x)$ a monotone function over $[0, \\infty)$.", + "Solution_1": "Solving the first , I get,\r\n\r\n$ x( (a^2 -1)x + 2a ) \\geq 0 $ Establishing anything beyond this requires whether |a| $>$ or $<$ 1\r\n\r\nwhen |a| $<$ 1 \r\nX $ \\in ( - \\infty , 0 ] \\cup [ \\frac{2a}{1- a^2) , \\infty ) }$\r\n\r\nThe roles are reversed when |a| > 1" +} +{ + "Tag": [ + "inequalities", + "geometry open", + "geometry" + ], + "Problem": "Given a triangle T, let P be a point such that it is on the boundary of T or inside of T, and let Q be a point at a distance d from P.\r\nLet r be the radius of the smallest circle with P as its center such that T is on the inside of this circle.\r\nLet R be the radius of the smallest circle with Q as its center such that T is on the inside of this circle.\r\n\r\nShow that $ r \\plus{} d\\leq 3R$.", + "Solution_1": "This one is almost trivial. Let $ A$ be the vertex of $ T$ at the distance $ r$ from $ P$. Since the circle centered at $ Q$ of radius $ R$ contains $ T$, it must contain both $ A$ and $ P$. Also $ |QA|\\ge |PA|\\minus{}d\\equal{}r\\minus{}d$. Thus, $ R\\ge \\max(r\\minus{}d,d)$ and we are left with showing that the inequality $ r\\plus{}d\\le 3\\max(r\\minus{}d,d)$ holds for all $ r,d>0$, which can be easily done by considering $ r>2d$ and $ r\\le 2d$ separately." +} +{ + "Tag": [ + "trigonometry" + ], + "Problem": "\u0388\u03c3\u03c4\u03c9 $ x,y,z \\in R^n$ \u03bc\u03b5 $ x \\neq y \\neq z \\neq x$ \u03ba\u03b1\u03b9 $ a_x \\equal{} \\angle (y \\minus{} x,z \\minus{} x),a_y \\equal{} \\angle (x \\minus{} y,z \\minus{} y),a_z \\equal{} \\angle (x \\minus{} z,y \\minus{} z)$.\u039d\u03b1 \u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9:\r\n1. $ a_x \\plus{} a_y \\plus{} a_z \\equal{} \\pi$\r\n\r\n2. $ \\frac {\\parallel{}x \\minus{} y\\parallel{}}{sina_z} \\equal{} \\frac {\\parallel{}y \\minus{} z\\parallel{}}{sina_x} \\equal{} \\frac {\\parallel{}z \\minus{} x\\parallel{}}{sina_y}$\r\n\r\n3. $ 1 < \\frac {(y \\minus{} x) \\cdot (z \\minus{} x)}{\\parallel{}y \\minus{} x\\parallel{} \\cdot \\parallel{}z \\minus{} x\\parallel{}} \\plus{} \\frac {(x \\minus{} y) \\cdot (z \\minus{} y)}{\\parallel{}x \\minus{} y\\parallel{} \\cdot \\parallel{}z \\minus{} y\\parallel{}} \\plus{} \\frac {(x \\minus{} z) \\cdot (y \\minus{} z)}{\\parallel{}x \\minus{} z\\parallel{} \\cdot \\parallel{}y \\minus{} z\\parallel{}} \\leq \\frac {3}{2}$\r\n\r\n4.$ 6 \\leq \\frac {\\parallel{}y \\minus{} x\\parallel{} \\cdot \\parallel{}z \\minus{} x\\parallel{}}{(y \\minus{} x) \\cdot (z \\minus{} x)} \\plus{} \\frac {\\parallel{}x \\minus{} y\\parallel{} \\cdot \\parallel{}z \\minus{} y\\parallel{}}{(x \\minus{} y) \\cdot (z \\minus{} y)} \\plus{} \\frac {\\parallel{}x \\minus{} z\\parallel{} \\cdot \\parallel{}y \\minus{} z\\parallel{}}{(x \\minus{} z) \\cdot (y \\minus{} z)}$ \u03b1\u03bd \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03c4\u03b1 $ x \\minus{} y,y \\minus{} z,z \\minus{} x$ \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03b1 \u03b1\u03bd\u03ac \u03b4\u03cd\u03bf.\r\n\r\n\u0391\u03bd \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd $ \\parallel{}x\\parallel{} \\equal{} \\parallel{}y\\parallel{} \\equal{} \\parallel{}z\\parallel{} \\equal{} R$ \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} 2$ \u03b5\u03af\u03bd\u03b1\u03b9 $ \\frac {\\parallel{}x \\minus{} y\\parallel{}}{sina_z} \\equal{} 2R$ \u03b5\u03bd\u03ce \u03b3\u03b9\u03b1 $ n>2$ \u03b5\u03af\u03bd\u03b1\u03b9 $ \\frac {\\parallel{}x \\minus{} y\\parallel{}}{sina_z} \\leq 2R$\r\n\u03a4\u03b1 1,2,3,4 \u03c4\u03b1 \u03ad\u03c7\u03c9 \u03ba\u03b1\u03c4\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03b1\u03bb\u03bb\u03ac \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03b3\u03b9\u03b1 $ n >2$ \u03cc\u03c7\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b7.\u0395\u03c5\u03c0\u03c1\u03cc\u03c3\u03b4\u03b5\u03ba\u03c4\u03b7 \u03b7 \u03cc\u03c0\u03bf\u03b9\u03b1 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1... :)", + "Solution_1": "[quote=\"giannis18\"]\u0388\u03c3\u03c4\u03c9 $ x,y,z \\in R^n$ \u03bc\u03b5 $ x \\neq y \\neq z \\neq x$ \u03ba\u03b1\u03b9 $ a_x \\equal{} \\angle (y \\minus{} x,z \\minus{} x),a_y \\equal{} \\angle (x \\minus{} y,z \\minus{} y),a_z \\equal{} \\angle (x \\minus{} z,y \\minus{} z)$.[/quote]\r\n\r\n:o\r\n\r\n\u0394\u03b5\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03c9... \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03c0\u03bf\u03c5 \u03c4\u03c1\u03af\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03bf\u03c1\u03af\u03b6\u03bf\u03c5\u03bd \u03ad\u03bd\u03b1 \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf, \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5 \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03b9\u03c2 \u03b1\u03c0\u03bb\u03bf\u03cd\u03c3\u03c4\u03b1\u03c4\u03b1 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac \u03c3\u03c4\u03bf 2-\u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03c0\u03bf\u03c5 \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b1 $ x,y,z$? :?:\r\n\r\n\u0389 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03c3\u03ba\u03bf\u03c0\u03cc\u03c2 \u03c4\u03b7\u03c2 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03b5\u03be\u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c3\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c5\u03c2 \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2? \u0395\u03bd\u03bd\u03bf\u03ce... \u03c0\u03ce\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03c0\u03c9, \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ce\u03c2 \u03c4\u03bf \u03b3\u03b5\u03b3\u03bf\u03bd\u03cc\u03c2 \u03cc\u03c4\u03b9 $ \\cos (60^\\circ) \\equal{} \\frac {1}{2}$ (\u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ae \u03b3\u03c9\u03bd\u03af\u03b1 and all that).\r\n\r\nDoes this help? :huh:\r\n\r\n\u03a3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7, \u03b1\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03bf\u03c0\u03c9\u03c3\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 (\u03bc\u03c0\u03c1\u03c1\u03c1\u03c1\u03c1...) \u03ba\u03b1\u03b9 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b2\u03b3\u03ac\u03bb\u03b5\u03b9 \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03b3\u03b9\u03b1 $ n \\equal{} 2$, \u03ad\u03c7\u03b5\u03b9\u03c2 \u03c4\u03b5\u03bb\u03b5\u03b9\u03ce\u03c3\u03b5\u03b9 \u03bc\u03cc\u03bb\u03b9\u03c2 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b1\u03b3\u03ae \u03b2\u03ac\u03c3\u03b7\u03c2 \u03c3\u03c4\u03bf \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03c0\u03bf\u03c5 \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b1 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 $ x \\minus{} y, y \\minus{} z$.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", + "Solution_2": "\u0391\u03c6\u03bf\u03cd \u03ad\u03c7\u03b5\u03b9\u03c2 $ \\parallel{}x\\parallel{}\\equal{}\\parallel{}y\\parallel{}\\equal{}\\parallel{}z\\parallel{}\\equal{}R$ \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03b3\u03b9\u03b1 $ n\\equal{}2$ \u03ba\u03b1\u03b9 $ n>2$.\u03a3\u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c4\u03bf R \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ad\u03c2 x,y,z \u03b5\u03bd\u03ce \u03c3\u03c4\u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c4\u03b1 x,y,z \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b5 n-\u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c4\u03b7 \u03c3\u03c6\u03b1\u03af\u03c1\u03b1 \u03bc\u03b5 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 R \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03ba\u03b1\u03c4'\u03b1\u03bd\u03ac\u03b3\u03ba\u03b7\u03bd \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03c0\u03bf\u03c5 \u03bf\u03c1\u03af\u03b6\u03bf\u03c5\u03bd \u03c4\u03b1 x,y,z.\u03a4\u03b1 \u03ac\u03bb\u03bb\u03b1 \u03b5\u03c1\u03c9\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1 \u03c4\u03b1 \u03ad\u03b2\u03b1\u03bb\u03b1 \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd \u03bc\u03ae\u03c0\u03c9\u03c2 \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c5 \u03b3\u03b9\u03b1 n>2", + "Solution_3": "[quote=\"giannis18\"]\u0391\u03c6\u03bf\u03cd \u03ad\u03c7\u03b5\u03b9\u03c2 $ \\parallel{}x\\parallel{} \\equal{} \\parallel{}y\\parallel{} \\equal{} \\parallel{}z\\parallel{} \\equal{} R$ \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03b3\u03b9\u03b1 $ n \\equal{} 2$ \u03ba\u03b1\u03b9 $ n > 2$.\u03a3\u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c4\u03bf R \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ad\u03c2 x,y,z \u03b5\u03bd\u03ce \u03c3\u03c4\u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c4\u03b1 x,y,z \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b5 n-\u03b4\u03b9\u03ac\u03c3\u03c4\u03b1\u03c4\u03b7 \u03c3\u03c6\u03b1\u03af\u03c1\u03b1 \u03bc\u03b5 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 R \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03ba\u03b1\u03c4'\u03b1\u03bd\u03ac\u03b3\u03ba\u03b7\u03bd \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03c0\u03bf\u03c5 \u03bf\u03c1\u03af\u03b6\u03bf\u03c5\u03bd \u03c4\u03b1 x,y,z.\u03a4\u03b1 \u03ac\u03bb\u03bb\u03b1 \u03b5\u03c1\u03c9\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1 \u03c4\u03b1 \u03ad\u03b2\u03b1\u03bb\u03b1 \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd \u03bc\u03ae\u03c0\u03c9\u03c2 \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c5 \u03b3\u03b9\u03b1 n>2[/quote]\r\n\r\n\u0388\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf, \u03b4\u03b9\u03ac\u03b2\u03b1\u03c3\u03b1 \u03c0\u03bf\u03bb\u03cd \u03b2\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac $ |x \\minus{} y| \\equal{} R$ \u03ba\u03bb\u03c0. \u03ba\u03b1\u03b9 \u03bf\u03cd\u03c4\u03b5 \u03ba\u03b1\u03bd \u03b5\u03af\u03b4\u03b1 \u03cc\u03c4\u03b9 \u03ae\u03b8\u03b5\u03bb\u03b5\u03c2 $ \\leq$ \u03b3\u03b9\u03b1 $ n > 2$. Sorry :oops:\r\n\r\nCheerio,\r\n\r\nDurandal 1707\r\n\r\nEdit: \u03b4\u03b5 \u03c3\u03bf\u03c5 \u03ad\u03b4\u03c9\u03c3\u03b1 \u03bb\u03cd\u03c3\u03b7!\r\n\u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 $ \\sin a_z \\equal{} \\frac {|(x \\minus{} z)\\times(y \\minus{} z)|}{|x \\minus{} z|\\cdot|y \\minus{} z|}$ \u03cc\u03c0\u03bf\u03c5 $ \\times$ \u03c4\u03bf \u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf. \u0386\u03c1\u03b1, \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03bc\u03ad\u03bb\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03ac\u03c2 \u03c3\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 (\u03c4\u03bf \u03b2\u03bb\u03ad\u03c0\u03b5\u03b9\u03c2 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03bc\u03b5 \u03ba\u03b1\u03b8\u03b1\u03c1\u03ae \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1 :P).\r\n\u03a4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03bc\u03ad\u03bb\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03c3\u03c6\u03ac\u03b9\u03c1\u03b1\u03c2 :)", + "Solution_4": "\u03a9\u03c7 \u03bd\u03b1\u03b9 \u03ae\u03c4\u03b1\u03bd \u03c0\u03bf\u03bb\u03cd \u03b1\u03c0\u03bb\u03cc \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac.\r\n\u0388\u03ba\u03b1\u03bd\u03b1 \u03cc\u03bc\u03c9\u03c2 \u03ad\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03bf 4. \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03c4\u03bf \u03b1\u03c0\u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac.(\u03c0\u03ae\u03c1\u03b1 \u03c4\u03bf 3. \u03c4\u03bf \u03b2'\u03bc\u03ad\u03bb\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c4\u03b9\u03ba\u03bf\u03cd-\u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03bf\u03cd \u03bc\u03ad\u03c3\u03bf\u03c5 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03ae \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bb\u03bf\u03b9 \u03bf\u03b9 \u03cc\u03c1\u03bf\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af).\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03bc\u03b5 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03af\u03be\u03c9;", + "Solution_5": "[quote=\"giannis18\"]\u03a9\u03c7 \u03bd\u03b1\u03b9 \u03ae\u03c4\u03b1\u03bd \u03c0\u03bf\u03bb\u03cd \u03b1\u03c0\u03bb\u03cc \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac.\n\u0388\u03ba\u03b1\u03bd\u03b1 \u03cc\u03bc\u03c9\u03c2 \u03ad\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03bf 4. \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03c4\u03bf \u03b1\u03c0\u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac.(\u03c0\u03ae\u03c1\u03b1 \u03c4\u03bf 3. \u03c4\u03bf \u03b2'\u03bc\u03ad\u03bb\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c4\u03b9\u03ba\u03bf\u03cd-\u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03bf\u03cd \u03bc\u03ad\u03c3\u03bf\u03c5 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03ae \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bb\u03bf\u03b9 \u03bf\u03b9 \u03cc\u03c1\u03bf\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af).\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03bc\u03b5 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03af\u03be\u03c9;[/quote]\r\n\r\n\u039a\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c1\u03c9\u03c4\u03ae\u03c3\u03c9: \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03bc\u03af\u03b1 \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 \u03ae \u03bc\u03af\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 $ \\frac {1}{\\cos A} \\plus{} \\frac {1}{\\cos B} \\plus{} \\frac {1}{\\cos C} \\geq 6$ \u03cc\u03c0\u03bf\u03c5 $ A,B,C$ \u03b3\u03c9\u03bd\u03af\u03b5\u03c2 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 (\u03ae \u03b1\u03bb\u03bb\u03b9\u03ce\u03c2 $ \\sum_{\\text{cyclic}}\\frac{bc}{a^2\\minus{}b^2\\minus{}c^2}\\geq3$ \u03b3\u03b9\u03b1 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ a,b,c$)?\r\n\r\n\u03a3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7, \u03b8\u03c5\u03bc\u03ae\u03c3\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03c3\u03b5 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03b5\u03c1\u03c9\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 WLOG: $ x \\equal{} 0, y \\equal{} (0,1), z \\equal{} (u,v)$ (\u03b1\u03bd \u03c3\u03bf\u03c5 \u03b1\u03c1\u03ad\u03c3\u03b5\u03b9 \u03c4\u03bf brute force, that is ;) ).\r\n\r\nCheerio,\r\n\r\nDurandal 1707", + "Solution_6": "\u038c\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03cc.\u0394\u03b5\u03bd \u03bc\u03b5 \u03c0\u03b5\u03b9\u03c1\u03ac\u03b6\u03b5\u03b9", + "Solution_7": "[quote=\"giannis18\"]\u038c\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03cc.\u0394\u03b5\u03bd \u03bc\u03b5 \u03c0\u03b5\u03b9\u03c1\u03ac\u03b6\u03b5\u03b9[/quote]\r\n\r\nActually, \u03b7 \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 \u03bf\u03be\u03c5\u03b3\u03ce\u03bd\u03b9\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 (\u03bf\u03c0\u03cc\u03c4\u03b5 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c7\u03b1\u03c1\u03ac \u03b1\u03c6\u03bf\u03cd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03b5\u03c3\u03c9\u03c4. \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03b1 $ \\geq 0$). \u039c\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03b1\u03bc\u03b2\u03bb\u03c5\u03b3\u03ce\u03bd\u03b9\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 \u03b1\u03bd \u03b8\u03ad\u03c3\u03b5\u03b9\u03c2: $ x\\equal{}\\minus{}1, y\\equal{}0, z\\equal{}1$.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", + "Solution_8": "\u0391\u03bd \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 $ x\\equal{}(\\minus{}1,\\minus{}1,...,\\minus{}1)$ \u03ba\u03c4\u03bb \u03b1\u03c5\u03c4\u03ac \u03b4\u03b5 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03bf\u03c5\u03bd \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03b1\u03c6\u03bf\u03cd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac. :maybe:", + "Solution_9": "[quote=\"giannis18\"]\u0391\u03bd \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 $ x \\equal{} ( \\minus{} 1, \\minus{} 1,..., \\minus{} 1)$ \u03ba\u03c4\u03bb \u03b1\u03c5\u03c4\u03ac \u03b4\u03b5 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03bf\u03c5\u03bd \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03b1\u03c6\u03bf\u03cd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac. :maybe:[/quote]\r\n\r\n\u0395\u03bd\u03bd\u03bf\u03ce \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ \\minus{} 1,0,1$ \u03c4\u03bf\u03c5 $ \\mathbb{R}$ \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03bc\u03b9\u03b1 \u03c7\u03b1\u03c1\u03ac \u03c4\u03b9\u03c2 \u03c0\u03c1\u03bf\u03cb\u03c0\u03bf\u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b7\u03c2 \u03ac\u03c3\u03ba\u03b7\u03c3\u03ae\u03c2 \u03c3\u03bf\u03c5 (\u03b3\u03b9\u03b1 $ n \\equal{} 1$), \u03b1\u03bb\u03bb\u03ac \u03cc\u03c7\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2. \u0394\u03b5 \u03c3\u03b5 \u03bd\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9 \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03ae \u03b1\u03bd \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03cc \u03c3\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03ba\u03c6\u03c5\u03bb\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf \u03ba\u03b1\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03af\u03b1 \u03b3\u03c9\u03bd\u03af\u03b1 \u03af\u03c3\u03b7 \u03bc\u03b5 \u03c0.\r\n\r\n\u0391\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03bf\u03c0\u03c9\u03c3\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03cc \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03b9\u03c2, \u03c0\u03ac\u03c1\u03b5 \u03b3\u03c9\u03bd\u03af\u03b5\u03c2 120, 30, 30 (\u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $ x \\equal{} (0,0)$, $ y \\equal{} (1,0)$, $ z \\equal{} \\frac {1}{2}( \\minus{} 1,\\sqrt {3})$).\r\n\r\nCheerio,\r\n\r\nDurandal 1707", + "Solution_10": "\u039f\u03ba \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1 :)" +} +{ + "Tag": [ + "percent", + "geometry", + "geometric transformation", + "probability", + "expected value" + ], + "Problem": "As the topic says, what is your IQ? Just wondering, because I would be willing to bet that the arithmetic mean value is pretty high around here...\r\n\r\nThis topic inspired by the \"Mensa International\" topic.", + "Solution_1": "I think there are enough topics open about this subject yet? :?", + "Solution_2": "AAAAAAAAA!!!!! NOT ANOTHER ONE!!!!!!! PLEASE!!!!!!! \r\n\r\nSeriously, this IQ thing's getting really annoying. Who cares if my iq is 100 or 180? Would it matter to you?", + "Solution_3": "My answer is: higher than that of any of the people who care about IQ. \r\n\r\nHowever, I respect people who self-educate about the facts about [url=http://learninfreedom.org/iqbooks.html]IQ testing and human intelligence[/url], so I reply to indicate that there is more to learn about that subject. \r\n\r\nP.S. What did you get on your last AMC test?", + "Solution_4": "Seriously, I don't really care about \"IQ\" or even believe in \"IQ\" tests. I'm just interested in the topic. The subject might as well be \"What's your SAT score?\", but IQ tests are specifically designed for what they measure (however accurate that measurement is really is). \r\n\r\np.s. I got a 118 on my last AMC-12.", + "Solution_5": "[quote=\"ZennyK\"]Seriously, I don't really care about \"IQ\" or even believe in \"IQ\" tests. I'm just interested in the topic. The subject might as well be \"What's your SAT score?\", but IQ tests are specifically designed for what they measure (however accurate that measurement is really is). \n\np.s. I got a 118 on my last AMC-12.[/quote]\r\n\r\nYou know, these subjects can really make people dejected. How would you feel if you saw that everone here had an iq of 140+ and yours was only 105? Not so hot, eh? That's one of the reasons why I don't like these topics about IQs. Sure you could say the same about the SATs or AMC/AIME/USAMO scores or MC score, but somehow, they're just different. Bah, I might be just putting off some of my ill feelings about IQ tests, because I think they're mainly garbage. Why do you want to know how smart you are? We've got tests and grades for that. Oh well, I'm probably boring everyone with my rant, so I'll stop it here.", + "Solution_6": "HAHAHAHA!!!!! Who said 200+ and who said 85<. IQ is stupid.", + "Solution_7": "My IQ is 80. So WHAT????????????????????????????????? Will you ever get to know me? Hehe. It doesn't really matter, since IQ can only get you so far in life.", + "Solution_8": "Quote from...erm...I'm not sure which of my classes has it posted on the wall but it's one of them...\r\n\"Your I WILL is more important than your I.Q.\"\r\nThe most ridiculous thing about IQ tests (for me at least) is the bonus points some of them give for speed. Speed will get you NOWHERE in real life, though I admit it's helpful on contests. Plus if you've seen problems before, and if you keep taking these tests you'll see just about any problem that's ever going to show up, you'll definitely score higher.", + "Solution_9": "Actually, speed, overall, decreases efficiency, but it is a measure of who can be MORE efficient in the same amount of time, or if given less time, will one maintain their efficiency. It also measures how fast you react to certain patterns, which often is very useful even in real life.", + "Solution_10": "is there a good online place that estimate your iq? i dont think iqtest.com and emode are that great..\r\n\r\nmy friend got 160 on iqtest.com and 95 on emode..", + "Solution_11": "I think most of them are all garbage. But there might be one that's semi-good out there.", + "Solution_12": "[quote=\"white_horse_king88\"]Actually, speed, overall, decreases efficiency, but it is a measure of who can be MORE efficient in the same amount of time, or if given less time, will one maintain their efficiency. It also measures how fast you react to certain patterns, which often is very useful even in real life.[/quote]\r\nOh, I know...I just dislike it because, well, because it lowers my IQ. LOL. It's also why I dislike the Countdown Round and various other things that I'll never be good at because I'm a comparatively slow reactor. Though my speed isn't bad enough to hurt me during contests like the AMC, so I don't actually have a right to complain, but...", + "Solution_13": "luckily, in america, success is based primarily on people-smarts, hard work, passion, and initiative. \r\n\r\nbut through the qualifying test for gifted and talented, i got a 147. you see, i figure that if i marry a man with an IQ of say, 200, our kids' IQ's will average out. that probably isn't true, but whatever.\r\n\r\nby the way, that was a joke. i'm not that shallow.. really.", + "Solution_14": "Good luck finding someone with a 200 IQ *pats*. That's almost as bad as bringing a tape measure on dates. (Don't worry if you didn't get what I just said. :-) )\r\n\r\nBut anyways...IQ shouldn't matter that much. It seems like the one IQ test I did overestimated my IQ, and since no one actually agrees on a \"fair\" or \"good\" IQ test, it's very likely that the scales are different anyways. After all, there's no standard.", + "Solution_15": "[quote=\"tokenadult\"]My answer is: higher than that of any of the people who care about IQ.[/quote] \n\nOnce again tokenadult has made a wonderful observation. :) Thank you.\n\nYours truly,\n\nM.E.B.\n\nBy the way, toekenadult's website is worth a look.\n\n[quote=\"toeknadult\"]However, I respect people who self-educate about the facts about [url=http://learninfreedom.org/iqbooks.html]IQ testing and human intelligence[/url], so I reply to indicate that there is more to learn about that subject.[/quote]", + "Solution_16": "[quote=\"jli\"]go to http://www.tickle.com\n\nI not sure if they are accurate though[/quote]\r\n\r\nNo, it's not very accurate at all. I've gone there and gotten a result of 183, which is probably far from true.", + "Solution_17": "[quote=\"M.E.B.\"][quote=\"tokenadult\"]My answer is: higher than that of any of the people who care about IQ.[/quote] \n\nOnce again tokenadult has made a wonderful observation. :) Thank you.\n\nYours truly,\n\nM.E.B.\n\nBy the way, toekenadult's website is worth a look.\n\n[quote=\"toeknadult\"]However, I respect people who self-educate about the facts about [url=http://learninfreedom.org/iqbooks.html]IQ testing and human intelligence[/url], so I reply to indicate that there is more to learn about that subject.[/quote][/quote]\r\nHurray for homeschooling! :)", + "Solution_18": "As by double post, one thread seems like it is going to be deleted, I wanted to save this quote from someone there that applies to this topic\r\n\r\n[quote=\"Annika\"]Sorry, but I have the feeling that people who care about IQ seriously misunderstood some things.\n\nAnyway, internet IQ tests are not reliable at all (they want to sell you your \"Complete Intelligence Profile\" or stuff like that, so what is their aim? To make you proud of your \"IQ 160\" or whatsoever so that you want to buy their certificate to show off with it).\nYou get better IQ tests (properly calibrated) from psychologist and at Mensa, but I'm conviced anyone can score really high in an IQ test if he or she does training for the question types.\n[/quote]I just think that she has a very good point ;)", + "Solution_19": "yeea... but what if i got the same score on internet as the administered ones? Am i truely intellgent as they sayy?? I mean I don't train with this kind of stuff, either.", + "Solution_20": "Does anyone have any idea of the IQ of people who actually design IQ tests? If it's like around 125 (arbitrary value), then what exactly would be the difference between someone with 139 IQ and 180 IQ? Wouldn't the difference be pretty much useless?", + "Solution_21": "[quote=\"Simfish\"]Does anyone have any idea of the IQ of people who actually design IQ tests? If it's like around 125 (arbitrary value), then what exactly would be the difference between someone with 139 IQ and 180 IQ? Wouldn't the difference be pretty much useless?[/quote]Again, the IQ measures just a few trainable skills, very specific. Then it is actually not that hard to make such differences. For example, I have never done any gymnastics, and even when I haven't ever study anything about it, I can tell the difference between a good and a very good gymnast. I suppose that if you have some experience on gymnastics at all, and an objective scale you can do a fair numerical scale on a group of gymnasts. It is even easier and more objective with the IQ.", + "Solution_22": "well, here's a question for you. has anyone seen the books at bookstores that claim to test your small child's IQ? many of these tests are for pre-school age children (that have not learned to read yet). the question i have is- how can you figure out IQ at that young of an age? now, i realize that IQ supposedly tests reasoning skills independent of KNOWLEDGE, but i must say, it would seem that it still takes time for reasoning skills to develop, and that these would be immeasurable at, say, two or three.\r\n\r\nbut then there is the idea that many of you here were probably particularly bright as youngsters. how is \"bright\" measured? is this the same as IQ, or is it simply the thirst for knowledge? how does this all tie into biological chemistry? have i stepped over the bounds of the round table forum? do i need to go to the biology forum to ask these questions? do we HAVE a philosophy forum?\r\n\r\n\r\ni think we need one. or perhaps thats what this forum is for.", + "Solution_23": "I agree--a philosophy forum would be awesome.\r\n\r\nAs for IQ, I think it is completely useless. Just look at all the tests out there--even if it were accurate in some form, you'd never know what scale to compare to. I think IQ is just an attempt to measure intelligence, but our notion of \"intelligence\" isn't really what's important, especially as a general thing. It seems to me that \"intelligence\" is equivalent or comparable to \"potential\" in our sense. It is more important to combine this \"intelligence\" thing with actual work/knowledge. In fact, I think in many ways we have extrapolated \"intelligence\"--it is most likely essentially tied with many other things which also matter. We think we can compare by intelligence, but really intelligence is also dedication, curiosity, and other things that form the potential of a person. Anyway, there isn't really a good way to test people, since invariably you will have to be biased towards a group with strong skills in a certain area.", + "Solution_24": "Yeah, good point there. I really don't think that IQ tests are an accurate representation of mental capability. \r\n\r\nOne example of this would be if somebody was having a bad day and couldn't concentrate. Everyone has difference performance rates at different times.", + "Solution_25": "[quote=\"Simfish\"]Does anyone have any idea of the IQ of people who actually design IQ tests? If it's like around 125 (arbitrary value), then what exactly would be the difference between someone with 139 IQ and 180 IQ? Wouldn't the difference be pretty much useless?[/quote]\r\n\r\nThat's quite an astute comment. Many high IQ scores are obtained by test-takers whose test-giver is very average, average enough to muff the administration rules of the test. \r\n\r\nhttp://denver.rockymountainnews.com/justin/3.shtml", + "Solution_26": "\"people who brag about their IQ are losers.\"\r\n\r\nStephen Hawking--New York Times Magazine sometime in 2005\r\n\r\n'nuff said.", + "Solution_27": "[url=http://www.nytimes.com/2004/12/12/magazine/12QUESTIONS.html?ex=1116216000&en=5943a5ca57c0d014&ei=5070]Interview with Stephen Hawking[/url]", + "Solution_28": "[quote=\"From the Hawking Interview\"]Life would be tragic if it weren't funny. [/quote]\r\n\r\nWow...what a great man. I really respect him.", + "Solution_29": "[quote=\"ankur87\"][quote=\"From the Hawking Interview\"]Life would be tragic if it weren't funny. [/quote]Wow...what a great man. I really respect him.[/quote]That is a very interesting comment, specially comming from someone whose life has been very tough! (according to the minibiography I read like six or seven years ago)." +} +{ + "Tag": [ + "ratio" + ], + "Problem": "Wht volume of hydrogen chloride gas is produced by the reaction of 40cubic cms of chlorine with hydrogen? (All gas volumes are measured at r.t.p).", + "Solution_1": "Hi Pat007. Welcome to this forum.\r\nThe relevant chemical equation is: $ Cl_{2}(g)\\plus{}H_{2}(g)\\longrightarrow 2HCl(g)$. What is the volume of hydrogen used? And what is the meaning of r.t.p.?", + "Solution_2": "Thanks for your response,\r\nr.t.p = room temperature and pressure.", + "Solution_3": "Ok, but what is the volume of hydrogen used? Or hydrogen is in excess?", + "Solution_4": "Only the vol of CL2 was given (40cm3). \r\nI did the following: \r\ncl2 + h2= 2hcl\r\n71 + 2 = 73\r\nratio: 1 = 73/71 \r\n 1g= 1.03g\r\n\r\n71g=1mol=24dm3\r\n1dm3 = 2.96g\r\n.040dm3 = .118g\r\n.118g of cl2 x 1.03g = 0.122g hcl\r\n73g hcl= 24dm3 then 0.122g = 24/73 x .122g= .040dm3. \r\n\r\nMy answer above is incorrect. Correct answer is 4.8dm3. What am I missing? Is there a simpler way to calculate this?", + "Solution_5": "[quote=\"Pat007\"]Correct answer is 4.8dm3.[/quote]\r\n\r\nThat is impossible. 40 cm3 of chlorine correspond to 0.0016 mol (considering ideal gas behaviour), and so the maximum quantity of HCl that can be obtained is 0.0032 mol (= 79 cm3). Are you sure there is no other data provided by the problem?", + "Solution_6": "actually, you're right. The book says 80dm3. How did you get 0016 mols?", + "Solution_7": "Just use the ideal gas law, PV = nRT, with P = 1 atm, V = 0.040 L, R = 0.0826 L atm K-1 mol-1, and T = 298 K.", + "Solution_8": "cool, thank you very much indeed!" +} +{ + "Tag": [ + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "there are 13 men, who come from k castles, sitting around a table. And 1 BA = BP. \r\n\r\nIF A = 2C then DC=DA and, as IP//AC, CP=AI. Since IP//AC then BA=BP.\r\n\r\n\r\nIf BA = BP then A = 2C.\r\n\r\nIf BA=BP then, by simmetry, 0$. Prove that the sequence $ (a_n)$ has infinitely\r\nmany numbers divisible by $ 2009$.", + "Solution_1": "[u][i]Solution :[/i][/u]\r\n\r\nWe have $ 40 \\equal{} 5.2^3$ and $ 2009 \\equal{} 41.7^2$ and it follows that $ \\gcd(40,2009) \\equal{} 1$ And Using Euler's Theorem we have $ 40^{\\phi(2009)}\\equiv 1 (\\mod2009) \\implies 40^{5.6.7.8}\\equiv 1(\\mod2009)$ Which means that $ 40^{k!} \\equiv 1(\\mod2009)$ For $ \\forall k\\ge 7$\r\n\r\nOn the other hand we can easly prove that $ a_n \\equal{} a \\plus{} \\sum_{k \\equal{} 1}^{k \\equal{} n}40^{k!}$\r\n\r\nand We know that there exists some $ m > 0$ for which $ 40^1 \\plus{} 40^{2!} \\plus{} ... \\plus{} 40^{6!}\\equiv m (\\mod 2009)$\r\n\r\nOr we have $ a_{2009l \\minus{} m \\minus{} a} \\equal{} a \\plus{} 40^1 \\plus{} 40^{2!} \\plus{} ... \\plus{} 40^{6!} \\plus{} \\sum_{k \\equal{} 7}^{k \\equal{} 2009l \\minus{} m \\minus{} a}40^{k!}\\equiv 0 (\\mod 2009)$ Where $ l$ is chosen such that $ 2009l \\minus{} m \\minus{} a > 0$\r\n\r\nTherefore there are an infinetly numbers Satisfying our condition" +} +{ + "Tag": [ + "AMC", + "USA(J)MO", + "USAMO", + "IMO", + "IMO 2009" + ], + "Problem": "Albania does only one and it will be 23 May. What about other countries?", + "Solution_1": "Serbia, only one, been on 12-13th April, 2 days x 3 problems, here are they http://www.mathlinks.ro/viewtopic.php?t=271835", + "Solution_2": "Portugal does 4 exams, 3 hours length each.", + "Solution_3": "Strictly speaking, in Brazil student usually take four TSTs. However, the selection process is a bit more convoluted. It also takes into account the National Olympiad, three or four problem sets (all sent via mail) and, minorly, age and experience.\r\n\r\nSo Brazilian students usually have to solve about 50 problems in our selection process.\r\n\r\nBut I believe other countries' students go through that, isn't it?", + "Solution_4": "Does anybody know how do they do in China because as I saw no chinese contestant has came 3 times in IMO, eventhought one wrote 2 perfect papers?!", + "Solution_5": "Greece does one exam-four problems", + "Solution_6": "In Argentina we have a single selection test, two days, three problems are given in each day, and contestants have 4 hours to solve the problems", + "Solution_7": "[quote=\"ridgers\"]Does anybody know how do they do in China because as I saw no chinese contestant has came 3 times in IMO, eventhought one wrote 2 perfect papers?![/quote]He maybe didn't have the age required for a 3rd participation. \r\n\r\nUsually in Romania there are anywhere from 5 to 8 TSTs (usually around 18-20 problems total), and before that there is the National Olympiad (you have to be in the top 15 of your grade to be able to take the tests initially - either that, or, to have won a medal at the IMO in the previous year.)", + "Solution_8": "Now China's TST consists of eight tests. Each of them contains 3 problems. six of them are \"quiz\" in which each problem worths 7 points and the other two are quite important because in which each problem worths 21 points.\r\n\r\nSince the population in China is very large, it is extremely hard to be in the IMO team for three times. In history there was only one guy who was in the team for three times.But in first year(1998), IMO was held in Taiwan and our team didn't go, so he only had two gold medals in IMO.", + "Solution_9": "in india as far as i know , there r 3 tests", + "Solution_10": "In Iran we had almost 10 exams this year ... each year they teach some of university books to us and take 4,5 exams from these lessons . 5 remaining exams consist our TST and our national olympiad . Iran TST held in 4 days and each day has 3 problem .", + "Solution_11": "In Turkey two days x 4 problems (mediterranean and silk road) and two days x 3 problems (Turkey TST)", + "Solution_12": "In Hongkong, we have one prelim contest, three tests, and a final APMO as the last selection, worthing 1, 2, 3, 3, 4 points respectively. So the total score is usually 13...\r\n\r\nusually, the first three tests are usually moderately easy, but I did poorly in the last two.\r\n\r\nWell, I got approx 6 only...", + "Solution_13": "In Ukraine 4 days x 3 problems x 12 contestants (to choose 6 of them)", + "Solution_14": "germany has 2 pre-selection and 7 selection exams.\r\n\r\nNaphthalin", + "Solution_15": "In Indonesia, after the national olympiad, the upcoming participant comes to the first phase of team selection camp. 5 tests are given in this phase. Then, 15 students were chosen to the second phase. 5 tests were also given in this phase, and also APMO.\r\nSo, I think Indonesia's got 10 tests. Pretty much, eh?", + "Solution_16": "U.S. has USAMO (6*7=42 points) + 3 TSTs (each consists of 3 questions, giving 9*7=63 points) = 105 points total", + "Solution_17": "[quote]U.S. has USAMO (6*7=42 points) + 3 TSTs (each consists of 3 questions, giving 9*7=63 points) = 105 points total[/quote]\r\nI wonder how many points person should have scored to get into USA IMO team? :)", + "Solution_18": "[quote=\"birzhan\"][quote]U.S. has USAMO (6*7=42 points) + 3 TSTs (each consists of 3 questions, giving 9*7=63 points) = 105 points total[/quote]\nI wonder how many points person should have scored to get into USA IMO team? :)[/quote]\r\n\r\nDepends on a year and several factors: generation of students, the questions given, etc.", + "Solution_19": "[quote=\"birzhan\"][quote]U.S. has USAMO (6*7=42 points) + 3 TSTs (each consists of 3 questions, giving 9*7=63 points) = 105 points total[/quote]\nI wonder how many points person should have scored to get into USA IMO team? :)[/quote]\r\n\r\nMy USAMO score this year is 31/42. I guess my TST score should be around 38/63 (failed the first day). So I probably got a 69/105 and made the last spot on the team.", + "Solution_20": "Azerbaijan had 5 team selection tests this year", + "Solution_21": "[quote=\"b555\"]in india as far as i know , there r 3 tests[/quote]\r\nYou are right :) \r\nWe have to get school recommendations to participate in the regional test; then they take the qualifiers to INMO (Indian national MO) and then there is a training camp (IMO TC) from where 6 are selected or IMO :)", + "Solution_22": "Don't forget the AMTI.", + "Solution_23": "[quote=\"nobelium\"]Don't forget the AMTI.[/quote]\r\nWell, here (in Kolkata) you do not need to give any pre-RMO tests. :)" +} +{ + "Tag": [ + "function", + "algebra unsolved", + "algebra" + ], + "Problem": "I [b]need[/b] a proof of following:\r\nLet $ Q(x)=(x-a)^{k}Q_{1}(x)({k}\\in{N}), Q_{1}(a)\\neq0, deg(P(x))1$ and $\\lfloor a^{m}\\rfloor \\not=\\lfloor b^{n}\\rfloor $ for any positive integers $m$ and $n$?", + "Solution_1": "Yes, take for example $ a \\equal{}\\sqrt{3}$ and $ b \\equal{}\\sqrt{6}$ and assume that $ \\lfloor a^{m}\\rfloor \\equal{}\\lfloor b^{n}\\rfloor \\equal{} k$ for some positive integers $ m,n,k$. Then we need $ k^{2}\\le 3^{m},6^{n}< (k\\plus{}1)^{2}$, so that $ |3^{m}\\minus{}6^{n}|\\le 2k$. Since obviously $ m > n$ we have that $ 2k\\ge 3^{n}.|3^{m\\minus{}n}\\minus{}2^{n}|\\ge 3^{n}$. But then $ 4.6^{n}\\le 4k^{2}\\le 9^{n}$, so $ 4\\le\\left(\\frac{3}{2}\\right)^{n}$, so $ n\\le 3$. But $ (\\lfloor b^{n}\\rfloor) \\equal{} 2,6,14,...$, while $ (\\lfloor a^{m}\\rfloor) \\equal{} 1,3,5,9,15,...$, so no terms are equal, QED.", + "Solution_2": "[quote=\"Peter\"]Yes, take for example $ a \\equal{} \\sqrt {3}$ and $ b \\equal{} \\sqrt {6}$ and assume that $ \\lfloor a^{m}\\rfloor \\equal{} \\lfloor b^{n}\\rfloor \\equal{} k$ for some positive integers $ m,n,k$. Then we need $ k^{2}\\le 3^{m},6^{n} < (k \\plus{} 1)^{2}$, so that $ |3^{m} \\minus{} 6^{n}|\\le 2k$. Since obviously $ m > n$ we have that $ 2k\\ge 3^{n}.|3^{m \\minus{} n} \\minus{} 2^{n}|\\ge 3^{n}$. But then $ 4.6^{n}\\le 4k^{2}\\le 9^{n}$, so $ 4\\le\\left(\\frac {3}{2}\\right)^{n}$, so $ n\\le 3$. But $ (\\lfloor b^{n}\\rfloor) \\equal{} 2,6,14,...$, while $ (\\lfloor a^{m}\\rfloor) \\equal{} 1,3,5,9,15,...$, so no terms are equal, QED.[/quote]\r\nSorry I don't quite catch for the statement $ k^{2}\\le 3^{m},6^{n} < (k \\plus{} 1)^{2}$, so that $ |3^{m} \\minus{} 6^{n}|\\le 2k$\r\nBut we can choose $ a \\equal{} 2 \\plus{} \\sqrt {3}$ $ b \\equal{} \\frac {5 \\plus{} \\sqrt17}{2}$ easily prove $ \\lfloor a^{m}\\rfloor$is odd and $ \\lfloor b^{n}\\rfloor$is even.Leads to contradiction." +} +{ + "Tag": [], + "Problem": "We were talking about research accessible to people at the intermediate level. Here is a problem to start us off. So far I've found 4 of the 5 Kaprekar numbers less than 100.\r\n\r\nAn $n$-digit positive integer $N$ is a [i]Kaprekar number[/i] if the sum of the number formed by the last $n$ digits in $N^2$, and the number formed by the first $n$ or $n-1$ digits in $N^2$ equals $N$. For example, $297$ is a Kaprekar number since $297^2=88209$ and $88+209=297$. Find all three-digit Kaprekar numbers.\r\n\r\nGood luck :lol:", + "Solution_1": "well its pretty easy to see that all numbers of the form 10^n -1 work ie: 9, 99, 999 ect.\r\n\r\nand then I found that if K is a n-digit kaprekar number then 10^n - K is a kaprekar number too. \r\n\r\nso using that I have 297, 703 and 999 for 3 digit kaprekar number, however I don't know how to find one yet (I just got 297 from you). \r\n\r\nfor 1 and 2 digits I found 1, 9, 45, 55, and 99 but again i just played around and found 45 which gave be 55 later." +} +{ + "Tag": [ + "geometry", + "integration", + "calculus", + "calculus computations" + ], + "Problem": "Find the area enlosed by the curve $ y \\equal{} x^{\\frac{3}{2}}$ and x-axis from $ x\\equal{}0$ to $ x\\equal{} 1$", + "Solution_1": "$ \\int_{0}^{1}x^{3/2}dx \\equal{} \\int_{0}^{1}(\\frac {2}{5}x^\\frac{5}{2})'dx \\equal{} \\frac {2}{5}$" +} +{ + "Tag": [ + "function", + "number theory unsolved", + "number theory" + ], + "Problem": "find all functions $ f: N^*\\to N^*$ such that $ f(mf(n)) \\equal{} n \\plus{} f(2004m)$\r\nI have a hint from my fiend,the result is not exist,but I can find the solution :(", + "Solution_1": "Let $ f(2004)\\equal{}c$. Put $ m\\equal{}1$, we have $ f(f(n))\\equal{}n\\plus{}c$. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2749]here[/url]. Hence $ f(n)\\equal{}n\\plus{}\\frac{c}2$ for all $ n\\in\\mathbb{N}^\\plus{}$ and $ c$ must be even. If we put $ m\\equal{}2,n\\equal{}1$ into the given equation, we have $ f(2f(1))\\equal{}1\\plus{}f(4008)$. Note that $ f(2f(1))\\equal{}2\\plus{}\\frac32c$ and $ 1\\plus{}f(4008)\\equal{}4009\\plus{}\\frac{c}2$. Hence $ c\\equal{}4007$ which is odd. We get a contradiction. Therefore, there is no such function.", + "Solution_2": "[quote=\"joh\"]Let $ f(2004) \\equal{} c$. Put $ m \\equal{} 1$, we have $ f(f(n)) \\equal{} n \\plus{} c$. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2749]here[/url]. Hence $ f(n) \\equal{} n \\plus{} \\frac {c}2$ for all $ n\\in\\mathbb{N}^ \\plus{}$ and $ c$ must be even. If we put $ m \\equal{} 2,n \\equal{} 1$ into the given equation, we have $ f(2f(1)) \\equal{} 1 \\plus{} f(4008)$. Note that $ f(2f(1)) \\equal{} 2 \\plus{} \\frac32c$ and $ 1 \\plus{} f(4008) \\equal{} 4009 \\plus{} \\frac {c}2$. Hence $ c \\equal{} 4007$ which is odd. We get a contradiction. Therefore, there is no such function.[/quote]\r\nI saw your \"here \" but I cant understand that step(I think it is also hardest one)\r\nCan you give me another hint! :)" +} +{ + "Tag": [], + "Problem": "Find the difference between the sum of all positive four digit palindromes and the sum of all positive three digit palindromes.\r\n\r\nThis is rather a challenge for beginners, but take a shot at it.", + "Solution_1": "u know \r\n\r\nit was one of problems from math chat sponsored by jwec05\r\n\r\nhe said it was foolish to add them all up arithmatically\r\n\r\nfirst u have to know what a palindrome is\r\n\r\nit is a word or group of words that is the same whether you read it forwards from the beginning or backwards from the end\r\n\r\nso it can be like..\r\n1001,1111,1221,1331,1441,1551,1661,1771,1881,1991,2002..\r\n\r\nfrom 1001 to 1991, it equals to ....\r\n\r\n(1001 x 10) + (00+11+22+33+44+55+66+77+88+ 99)\r\n\r\nadn therefore u can get the answer easily", + "Solution_2": "my brother just solved this problem in a much easier way\r\n\r\nsince it's a palindrome u can state 4 digiter like this\r\n\r\n(A x 1000) + (B x 100) + (B x 10) + (A)\r\n\r\nu can state 3 digiter palindrome like this\r\n\r\n(A x 100) + (B x 10) + (A)\r\n\r\nA can be one of these numbers..\r\n1 ~ 9 so the sigma of it is 45\r\nB can be one of these numbers..\r\n0 ~ 9 so it also has the sigma of 45\r\n\r\nso restate the phrase like his\r\n4 digiter:\r\n(45 x 1000) + (45 x 100) + (45 x 10) + (45)\r\n3 digiter:\r\n(45 x 100) + (45 x 10) + (45)\r\n\r\nsince we are supposed to find the difference of them, u subtract these.\r\nwhich u would get\r\n(45 x 900) + (100 x 45)\r\nwhich is also\r\n40500 + 4500\r\nand therefore u get\r\n45000\r\n\r\ngod I'm so lucky to have such a smart brother", + "Solution_3": "my brother made a stupid mistake guys\r\n\r\nhe forgot to multiplay 9 and 10's\r\n\r\nBaseballJoe2002 fixed my brother's solution\r\n\r\n\r\nPiratevsPirate: Find the difference between the sum of all positive four digit palindromes and the sum of all positive three \r\ndigit palindromes. \r\nPiratevsPirate: since it's a palindrome u can state 4 digiter like this\r\n\r\n(A x 1000) + (B x 100) + (B x 10) + (A)\r\n\r\nu can state 3 digiter palindrome like this\r\n\r\n(A x 100) + (B x 10) + (A)\r\n\r\nA can be one of these numbers..\r\n1 ~ 9 so the sigma of it is 45\r\nB can be one of these numbers..\r\n0 ~ 9 so it also has the sigma of 45\r\n\r\nso restate the phrase like his\r\n4 digiter:\r\n(45 x 1000) + (45 x 100) + (45 x 10) + (45)\r\n3 digiter:\r\n(45 x 100) + (45 x 10) + (45)\r\n\r\nsince we are supposed to find the difference of them, u subtract these.\r\nwhich u would get\r\n(45 x 900) + (100 x 45)\r\nwhich is also\r\n40500 + 4500\r\nand therefore u get\r\n45000\r\n\r\nPiratevsPirate: that's my brohter's solution\r\nPiratevsPirate: what do u think?\r\nBaseballJoe2002: hmm...i m not quite sure about his method\r\nPiratevsPirate: what's ur solution then?\r\nPiratevsPirate: but is the answer correct?\r\nBaseballJoe2002: i think the four digit ones should be 10*45*100+9*45*100+9*45*10+10*45\r\nPiratevsPirate: is * a multiplication?\r\nBaseballJoe2002: yes\r\nBaseballJoe2002: o..that should be 10*45*1000 for the first one\r\nPiratevsPirate: y do u do that?\r\nBaseballJoe2002: ok...let's suppose that the first digit was a one...then the last digit would have to be a one\r\nPiratevsPirate: yeah....\r\nBaseballJoe2002: we would have 1_ _ 1 and there are ten possibilities for the middle two digitss\r\nPiratevsPirate: I get it now!!\r\nPiratevsPirate: hold on...\r\nPiratevsPirate: then u r right\r\nBaseballJoe2002: ok\r\nPiratevsPirate: then is the restated phrase of 3 digits wrong also?\r\nBaseballJoe2002: yeah\r\nPiratevsPirate: wait\r\nPiratevsPirate: but the number he wrote was like this\r\nPiratevsPirate: 1000 x A which is the 4th digit / 100 x B is the 3rd digit / 10 x B is the 4th digit / A\r\nPiratevsPirate: that's right isn't it?\r\nBaseballJoe2002: yes\r\nPiratevsPirate: I don't think u have to multiply 10 on the fourth and the first \r\nBaseballJoe2002: both your brother and I used that same reasoning...i believe he just forgot to factor in how many times each number can occur\r\nPiratevsPirate: digits\r\nBaseballJoe2002: let's go back to a=1...then we have 1 _ _ 1. the middle two numbers have 10 choices. So then if we considering just how much taht first 1 contributes to the sum we find that it contributes 1000 for every time it is part of a number. It is part of 10 numbers.\r\nBaseballJoe2002: so we have 10*1000\r\nBaseballJoe2002: you can do that with every value a can be\r\nBaseballJoe2002: e.g. if a=2 we have 10*2000, if a=3 then 10*3000\r\nBaseballJoe2002: etc.\r\nBaseballJoe2002: and the sum for the those numbers will be 10*1000 + 10*2000 +...+ 10*9000 =10*45*1000\r\nPiratevsPirate: okay...\r\nBaseballJoe2002: you can apply the same reason for the hundreds, tens and units place\r\n\r\nso my solution is wrong and joe is right\r\n\r\nthe answer is supposed to be like this\r\nI will right on the next page", + "Solution_4": "the 4 digit palindrome :\r\n\r\n(10x45x1000) + (9x45x100) + (9x45x10) + (10x45)\r\n ^ ^ ^ ^\r\n these numbers were supposed to be multiplied because in the number\r\n\r\n1--1 there are 10 possiblilites for the -- to be filled\r\n\r\nso u multiply 10 for the 1st and the 4th digit, and 9 for 2nd and 3rd digits \r\n(9 is for the outer number which has only 9 probabilities)\r\n\r\nthe 3 digit palindrome :\r\n\r\n(10x45x100) + (9x45x10) + (10x45x1)\r\n\r\nthe difference would be like\r\n(10x45x900) + (9x45x10) which u would get\r\n\r\n= 409050", + "Solution_5": "First, how many 4-digit palindromes are there?\r\n\r\nFor the 1000's place, you can put any # from 1-9.\r\nFor the 100's place, you can put any # from 0-9.\r\nAfter that, the 10's and the 1's place are determined.\r\n\r\nso, 10 x 9 = 90 possible #'s.\r\n\r\nIn the 1000's place, of those 90, 10 are each #.\r\n so 10(1+2+ ... +8+9) * 1001.\r\n = 450 * 1001\r\n = 450450\r\n\r\nIn the 100's place, of those 90, 9 are each #.\r\n so 9(0+1+ ... +8+9) * 110.\r\n = 9(45) * 110\r\n = 405 * 110\r\n = 44550.\r\n\r\nTOTAL FOR FOUR DIGITS: 495000\r\n\r\n\r\nNext, how many 3-digit palindromes are there?\r\n\r\nFor the 100's place, you can put any # from 1-9.\r\nFor the 10's place, you can put any # from 0-9.\r\nAfter that, the 1's place is determined.\r\n\r\nso, 10 x 9 = 90 possible #'s.\r\n\r\nIn the 100's place, of those 90, 10 are each #.\r\n so 10(1+2+ ... +8+9) * 101.\r\n = 450 * 101\r\n = 45450\r\n\r\nIn the 10's place, of those 90, 9 are each #.\r\n so 9(0+1+ ... +8+9) * 10.\r\n = 9(45) * 10\r\n = 405 * 10\r\n = 4050.\r\n\r\nTOTAL FOR THREE DIGITS: 49500.\r\n\r\nDIFFERENCE BETWEEN THE TWO TOTALS: [b]445500[/b].", + "Solution_6": "Just an arithmatic problem....\r\n\r\nI think\r\n\r\nJoe and My brother checked it for me...\r\n\r\nhee hee hee\r\n\r\nmerry christmas!!!", + "Solution_7": "Happy Christmas, shinwoo and white_horse_king88!\r\n\r\n(I'm pretending to be British. :) )", + "Solution_8": "Happy New Year Pokey and white_horse_king88" +} +{ + "Tag": [ + "combinatorics open", + "combinatorics" + ], + "Problem": "let $g$ be the graph with vertices $v_1,...,v_p$.when it has a cycle with lenght of $p$?", + "Solution_1": "You question is bit ill put.\r\nHaving no cycle means being a tree (a definition also). \r\nHaving a cycle of size p is not very relevant structurally in graph theory.\r\nProperty would be less stringent like : \r\n- Having at least a cycle of size p\r\n- Having at least a cycle of size al least p", + "Solution_2": "In this case it's very relevant structurally because $p$ is also the number of vertices of the graph. Alas, the question is not easy at all, and is (I believe) NP-complete, meaning it's very hard to check even once you know what the graph is.", + "Solution_3": "if $\\delta \\ge 2$ we have a cycle with length of $p$" +} +{ + "Tag": [ + "vector", + "trigonometry" + ], + "Problem": "Given only the definition that the dot product of two vectors $\\vec{A}, \\vec{B}\\in \\mathbb{R}^{n}$ is given by\r\n\r\n$| \\vec{A}| | \\vec{B}| \\cos \\theta$\r\n\r\nWhere $\\theta$ is the angle between them, prove that the dot product is distributive over vector addition.", + "Solution_1": "Let $\\vec{A}= \\left[\\begin{array}{c}a_{1}\\\\\\vdots\\\\ a_{n}\\end{array}\\right]$ and $\\vec{B}= \\left[\\begin{array}{c}b_{1}\\\\\\vdots\\\\ b_{n}\\end{array}\\right]$. It of course sufficies to prove that $\\vec{A}\\cdot\\vec{B}= \\sum_{i=1}^{n}a_{i}b_{i}$. \r\nBy the cosine law, we have \r\n$\\left|\\vec{A}-\\vec{B}\\right|^{2}= \\left|\\vec{A}\\right|^{2}+\\left|\\vec{B}\\right|^{2}-2\\left|\\vec{A}\\right|\\cdot \\left|\\vec{B}\\right|\\cdot \\cos\\theta$. \r\nThus \r\n$\\vec{A}\\cdot \\vec{B}= \\frac{\\left|\\vec{A}\\right|^{2}+\\left|\\vec{B}\\right|^{2}-\\left|\\vec{A}-\\vec{B}\\right|^{2}}{2}= \\frac{\\sum_{i=1}^{n}a_{i}^{2}+b_{i}^{2}-\\left(a_{i}^{2}-2a_{i}b_{i}+b_{i}^{2}\\right)}{2}= \\sum_{i=1}^{n}a_{i}b_{i}$\r\nwhich ends the proof." +} +{ + "Tag": [ + "percent" + ], + "Problem": "Heh... After almost 7 months of working on my mersenne prime (the total thing), my hard drive has crashed and the work has gone down the drain! Its ok though, because AoPrimeS still was credited for the time my computer was working.\r\n\r\nWith all the new members coming in, I wanted to offer them to join AoPS's Mersenne Prime team (about 10 people so far). Basically, you download a program at mersenne.org called Prime 95 and it tries to find the next Mersenne prime. If you do download the program type in the following to join our group:\r\n\r\nUsername: AoPrimeS\r\nPassword: lovemath\r\n\r\n\r\n[url]http://www.mersenne.org[/url]", + "Solution_1": "what is a mersenne prime? plz explain -_-, sry for my ignorance", + "Solution_2": "A mersenne prime is a prime of the form [tex]2^p-1[/tex], where p is a prime.", + "Solution_3": "Right. 2^n-1. Recently, a guy found a huge mersenne prime (the 41st) They are looking for the 10 million digit prime and rewarding $100,000 to the first person who finds it. BTW, Prime 95 runs in the background of your computer and you hardly notice it's there. Without Prime 95, your computer basically has a whole bunch of unused memory, but with the program, it utilizes the unused memory to factor, and LL test the primes.", + "Solution_4": "I'm currently running MacLucas (another mersenne prime thing that there was a link to on http://www.mersenne.org, but it can't do the team thing) on my iMac, and Prime95 on my ultra-slow laptop.", + "Solution_5": "Cooool. Another new member! What would be really cool is if we could get just one school to all run computers. I'd do it with my school but we have bess.... I hate that dog.", + "Solution_6": "I'd love to do it and my computer has a ton of memory and stuff on it being a dell and I only use a small portion but my parents think it will get a virus on my comp so I can't do it.", + "Solution_7": "I think mine is about 70% (probably wrong, because I haven't use the other the Dell in a while). What happens when it reaches 100%.", + "Solution_8": "I'm still trying to figure out how to get the program on the mac to save. :(", + "Solution_9": "How long does it take for a computer to test if a number is prime? I left my computer on for about 10 hours and it did 0.16% of the test.\r\n\r\n(it's doing the Lucas test)", + "Solution_10": "it takes quite a while. you can get an estimate if you fiddle around with the controls...\r\n\r\nthe estimate will be quite early if you dont have your computer on 24/7 though...\r\n\r\nmine have been taking two to three months.", + "Solution_11": "two to three months? Are you doing the entire factoring test???\r\n\r\nI was doing it for nearly 6 months and it was only at like 70% :-P. BTW, I believe this program is virus free as a couple thousand people are using it. I haven't experienced any problems plus they give you the source code if you want to look at it.", + "Solution_12": "i started a factoring test (the big one...lucas-leihmer or something like that) just yesterday and its about 1% complete. My computer estimates completion in late july, but it will probably end up being a couple weeks to a month later...\r\n\r\nso yeah. two to three months. it really depends on how much you shut off your computer.", + "Solution_13": "Is LL the 10,000,000 digit request thing? Because I have a fast computer, and it took a long time. \r\n\r\nP.S. The Status thing is off by about half a year =D", + "Solution_14": "How much memory are you supposed to use? I'm using about 800 and it says that I'm at 50 percent after only two weeks", + "Solution_15": "800?? Wow that's a lot... I used 200 and it took me like 5 months =D. Use as much as you like.", + "Solution_16": "Wow... after another 2-3 months, I have finished my LL test! I guess it wasn't a prime :( ... Oh well. If anyone else wants to join now that we have tons of members, please do so! \r\n\r\nFun thing to do with the useless memory of your comp!\r\n\r\nAoPrimeS\r\npassword: lovemath", + "Solution_17": "[quote=\"Ragingg\"]Cooool. Another new member! What would be really cool is if we could get just one school to all run computers. I'd do it with my school but we have bess.... I hate that dog.[/quote]\r\n\r\nThe school my UMTYMP math teacher work at did exactly that. And he asked veryone in our math class to also do that. :)", + "Solution_18": "[quote=\"ragingg\"]What would be really cool is if we could get just one school to all run computers.[/quote]\r\n\r\nI suggested that to my math team coach, but he said the head computer guy probably wouldn't let us, even if it meant possibly winning 100 grand for our school..." +} +{ + "Tag": [ + "number theory proposed", + "number theory" + ], + "Problem": "[b]- Let $ q > 3$ be a prime number and $ k \\in \\mathbb{N}$.\n\nShow that we can always find $ a_{1}, a_{2}, \\ldots, a_{K} \\in \\mathbb{N}$ such that\n\n$ \\left\\vert \\{a_{1}, a_{2}, \\ldots, a_{K} \\} \\cap \\{3, 4, \\ldots \\} \\right\\vert \\geq 1$\n\nand\n\n$ 2^{k} \\equiv a_{1}!a_{2}!\\cdot \\ldots \\cdot a_{K}! \\mod q.$\n\nHope you all like it.\n\nI look forward to seeing your solutions soon![/b] :D", + "Solution_1": "Nice going, [b]coquitao[/b]!\r\n\r\nYour problem proposal is really cool! Keep up the good work, [i]hombre[/i]! :lol:" +} +{ + "Tag": [ + "integration" + ], + "Problem": "We give $\\omega_{0}$ inital angular velocity to a disc of radius $R$ and put it on the floor perpendicularly where the coefficient of kinetic friction is $\\mu$. When will it stop spinning?", + "Solution_1": "$\\mu mgR=I\\alpha$\r\n$I=\\frac{mR^{2}}{2}$\r\n$\\alpha=\\frac{2\\mu g}{R}$\r\n$w_{0}=\\alpha t$\r\n$t=\\frac{w_{0}R}{2\\mu g}$", + "Solution_2": "nope, thats not right. :( \r\nIf it would be that easy I wouldnt have posted it here. :wink:", + "Solution_3": "What does this mean?\r\n'We put it perpendicular.'\r\nDoesn't it lie on the floor?", + "Solution_4": "it's Namiqfizik's answer *3/2,right?you should use the cinematic of the rigid solid", + "Solution_5": "I think you need to assume that $\\mu$ is big enough, or else it might never stop. I'll try to post a full solution soon...", + "Solution_6": "here is the solution:\r\nThe torque of the friction force:\r\n$M=-\\int_{0}^{2\\pi}\\int_{0}^{R}\\frac{\\mu Mg}{R^{2}\\pi}r^{2}dr d\\varphi=-\\frac23 \\mu MgR$\r\nso the angular accelaration:\r\n$\\ddot \\varphi=\\frac{M}{\\Theta}=\\frac{M}{\\frac12 MR^{2}}=-\\frac43 \\frac{\\mu g}{R}$\r\nSo:\r\n$\\int_{\\omega_{0}}^{0}d\\dot \\varphi=\\int_{0}^{T}\\ddot \\varphi dt$\r\nFrom here the time when it stops spinning:\r\n$T=\\frac{3\\omega_{0}R}{4\\mu g}$", + "Solution_7": "Haha, now I understand, I've done this before. I was confused by your use of the word \"perpendicular.\" It made it seem as though the disc was standing on edge." +} +{ + "Tag": [ + "function", + "algebra unsolved", + "algebra" + ], + "Problem": "Prove that if [i]f[/i] is a real-valued functionm such that for all real x, f(x+1) + f(x-1) = 3^0.5 f(x), then f is periodic\r\nWhat is the smallest p ( p >0), such that f(x+p) = f(x) for all x?\r\n\r\nFrank Morgan", + "Solution_1": "$f(x+1) + f(x-1) = \\sqrt{3} f(x)$ (*)\r\n\r\nPlugging $x+1$ and $x-1$ in (*), summing, simplifying gives : \r\n\r\n$f(x-2) + f(x+2) = f(x)$ (**)\r\n\r\nPlugging $x-2$ in (**) and summing gives $f(x-4) + f(x+2) = 0$\r\n\r\nThen it's finished." +} +{ + "Tag": [ + "inequalities", + "geometry", + "inequalities unsolved" + ], + "Problem": "Let $\\ ABC$ a triangle whith side lengths $\\ a,b,c$ and area $\\ S$.Consider on the sides $\\ BC,CA,AB$ points \r\n$\\ A_1,B_1,C_1$ and let $\\ a_1,b_1,c_1$ the side lengths of the triangle $\\ A_1B_1C_1$ and $\\ S'$ its area.Prove that\r\n$\\ (b^2+c^2-a^2)a_1^2+\\ (c^2+a^2-b^2)b_1^2+\\ (a^2+b^2-c^2)c_1^2$>$\\ 8S(S-2S')$.When does the equality occur?", + "Solution_1": "Please, try it!I don't know to solve it and I think it's difficult!" +} +{ + "Tag": [], + "Problem": "Here is a poll that includes 58 options for who will be the winning team in 2007.\r\n\r\nActually, I cannot submit 58 options, so the last 8 were deleted.\r\n\r\nWow, I have to delete a lot of the options because of the poll limit :( \r\n\r\nNow, we know what the poll limit is :D \r\n\r\nI think I spelled everything correctly...", + "Solution_1": "Hah. FL has all of the votes :D.", + "Solution_2": "If my computer is serving me correctly, I believe there are not 50 choices in the poll...", + "Solution_3": "Once again I voted for Michigan just because that is my home state.", + "Solution_4": "[quote=\"pgpatel\"]Hah. FL has all of the votes :D.[/quote]\r\n\r\nHaha, no you don't.", + "Solution_5": "Isn't this thread just like the other one? Why do we need two? :? One is bad enough, jk :P", + "Solution_6": "Preetty much.\r\n\r\nIt's just that one author \"attempted\" to do the impossible. :lol:", + "Solution_7": "I give it a week before one of them is deleted by a mod.", + "Solution_8": "[quote=\"SplashD\"]I give it a week before one of them is deleted by a mod.[/quote]\r\n\r\nI give it 4 days and 13 hours.", + "Solution_9": "[quote=\"Ignite168\"][quote=\"SplashD\"]I give it a week before one of them is deleted by a mod.[/quote]\n\nI give it 4 days and 13 hours.[/quote]\r\n\r\nI smell a bet.", + "Solution_10": "Well, I tried to put 58 options in the poll, but evidently, you can only have 10 options in a poll, so I had to delete a bunch of them.", + "Solution_11": "[quote=\"b-flat\"]Well, I tried to put 58 options in the poll, but evidently, you can only have 10 options in a poll, so I had to delete a bunch of them.[/quote]\r\n\r\nMaybe you can put 5 states in one option.", + "Solution_12": "I'd give it 5 hours." +} +{ + "Tag": [ + "analytic geometry", + "graphing lines", + "slope" + ], + "Problem": "Find the equation of the tangent line of the curve y=x^2-4x-5 at (-2,7).", + "Solution_1": "Using diffrentials\r\n$ y'\\equal{}2x\\minus{}4$\r\nplugging, we obtain that at $ \\minus{}2\\equal{}x$\r\nthe slope is $ \\minus{}8$\r\n\r\nIm starting to worder, is this your comprehensive homework, and you are making uus do it?" +} +{ + "Tag": [ + "complex analysis", + "function", + "complex analysis unsolved" + ], + "Problem": "Let $f: \\mathbb C^2\\to\\mathbb C$ be an entire function such that $f(z_1+1,z_2)=f(z_1,z_2+1)=f(z_1+i,z_2+\\lambda i)=f(z_1,z_2)$ for all $z_1,z_2$. $\\lambda$ is an irational real number in $(0,1)$.\r\n\r\nProve that $f$ is constant.", + "Solution_1": "just a question: what is an entire function from $\\mathbb{C}^2$ to $\\mathbb{C}$ ?", + "Solution_2": "I mean holomorphic, which means $\\mathbb C$-differentiable." +} +{ + "Tag": [ + "vector", + "algebra unsolved", + "algebra" + ], + "Problem": "Consider a set $ S$ of $ n$, $ n \\ge 5$, triples $ (x_i,y_i,z_i)$ of real numbers, $ i \\equal{} 1, ..., n$, such that for all set $ A$ $ \\subset S$, $ |A|\\equal{}4$, there exists a triple of real numbers $ (a,b,c)$ such that $ ax_j\\plus{}by_j\\plus{}cz_j > 0$ for all $ (x_j,y_j,z_j) \\in A$. Does there exist a triple $ (d,e,f)$ of real numbers such that $ dx_i\\plus{}ey_i\\plus{}fz_i > 0$ for all $ (x_i,y_i,z_i) \\in S$?", + "Solution_1": "We can rephrase the problem if we think at the dot product: given $ n$ vectors in space, with the same origin $ O$, if any 4 lie in the same halfspace (passing through $ O$), then all lie in the same halfspace (also passing through $ O$.)", + "Solution_2": "Enescu,\r\n\r\nDo you if this is true? Is this a known result?", + "Solution_3": "I think we can use [url=http://en.wikipedia.org/wiki/Helly%27s_theorem]Helly's theorem[/url] to prove this. Let $ T_i$ be the set of points $ (a, b, c)$ such that $ ax_i \\plus{} b y_i \\plus{} c z_i > 0$. Each $ T_i$ is a convex region in 3-dimensional space. The given condition says that every 4 of the $ T_i$ intersect. So, by Helly's theorem, all of the $ T_i$ intersect. The common intersection point gives us the desired point $ (d, e, f)$." +} +{ + "Tag": [ + "arithmetic sequence" + ], + "Problem": "Consider a set of distinct 3-digit numbers each with this property: the sum of its digits is a 2-digit number, and the sum of THAT number's digits is a 1-digit number. How many elements are in the set, and what is the sum of those elements?", + "Solution_1": "Since it seems no one has a solution, let me try part of it. Actually I'm not good at this type of problem, so correct me if I'm wrong.\r\n\r\n[hide=\"number of elements\"]\nLet $S(n)$ represent the sum of the digits of $n$. Obviously if $n$ is a 3-digit number, then $1\\le S(n)\\le 27$. In addition, in order for $S(S(n))$ to be a 1-digit number (given that $S(n)$ is a 2-digit number), we must have $S(n)=10,11,12,13,14,15,16,17,18,20,21,22,23,24,25,26,27$\nIn other words, to find the number of 3-digit numbers $n$ to satisfy this condition, we can subtract the number of $n$ such that $S(n)=1,2,3,4,5,6,7,8,9,19$ from the number of all 3-digit numbers.\n\n$S(n)=1\\rightarrow n=\\underbrace{100}_{1}$\n$S(n)=2\\rightarrow n=\\underbrace{200}_{1},\\underbrace{110,101}_{2}$\n$S(n)=3\\rightarrow n=\\underbrace{300}_{1},\\underbrace{210,201}_{2},\\underbrace{120,111,102}_{3}$\n$\\vdots$\n$\\vdots$\n$S(n)=9\\rightarrow n=\\underbrace{900}_{1},\\underbrace{810,801}_{2},\\underbrace{720,711,702}_{3}, ...\\underbrace{...108}_{9}$\n\nAlso,\n$S(n)=19\\rightarrow n=\\underbrace{991,982,...919}_{9},\\underbrace {892,883,...829}_{8}, ...\\underbrace{199}_{1}$\n\nSo the total number of elements is\n$900-(1+(1+2)+(1+2+3)+...+(1+2+...+10)+(9+8+...+1))=900-265=\\boxed{635}$\n\nFor the sum, you can probably use the fact that each small set that I made (ex.$720,711,702$ etc.) is an arithmetic sequence with a common difference $9$, but I'm not sure.[/hide]\r\nHope I counted them correctly...", + "Solution_2": "i found a sum formula for them.\r\n$\\boxed{S_n=(((the highest number+the smallest number)-27*(n-1))/2)*n*(n+1)/2}$\r\ne.g.\r\n$S(1)=100$ and $S_1=(((100+100)-27*(1-1))/2)*1*2/2=100$\r\n$S(2)=200,101,110$ the sum of them is $200+101+110=411$\r\n$S_2=(((200+101)-27*(2-1))/2)*2*3/2=411$\r\n$S(3)=300,210,201,111,120,102$ the sum of them is $300+210+201+111+120+102=1044$\r\n$S_3=(((300+102)-27*(3-1))/2)*3*4/2=1044$\r\n$S(4)=400,301,310,220,202,211,130,103,112,121$ teh smu of them is $2110$\r\n$S_4=(((400+103)-27*(4-1))/2)*4*5/2=2110$\r\nbut for 19 i can't do with this formula.it is from 1 to 9.\r\nfrom 10 to 19 \r\n$\\boxed{S_n=S_9+4995*(n-9)-211*(n-10)*(n-9)*(n-8)/6-11*(n-11)*(n-10)*(n-9)/6}$\r\ni made these formulas by myself and i haven't got a proof.\r\nwith these we can calculate $S_{1,2...9}$ and $S_{19}$\r\n$S_1+S_2+S_3+S_4+....+S_9=100+411+1044+2110+3720+5985+9016+12924+17820=53130$\r\n$S_{19}=31635$\r\n$53130+31635=84765$\r\nwe will subtract it from the sum of all 3 digit numbers\r\n$494550-84765=409785$" +} +{ + "Tag": [ + "real analysis", + "calculus", + "integration", + "ratio", + "function", + "geometric series", + "absolute value" + ], + "Problem": "If $ f(x) \\equal{} \\frac {xe^{ \\minus{} 2x}}{1 \\minus{} e^{ \\minus{} 3x}}$ , then\r\n\\[ \\int\\limits_0^\\infty f(x) \\equal{} \\sum\\limits_{n \\equal{} 1}^\\infty \\frac {1}{(2 \\plus{} 3n)^2}\\]", + "Solution_1": "Not so strange. This is all very normal. Some hints:\r\n\r\n1. Geometric series!\r\n\r\n2. Whenever you see the integral of a sum, or the sum of integrals, or a double sum, or a double integral, [i]of course[/i] you want to interchange the order of operations. It's what you nearly always do. The Lebesgue theory is used to provide justifications for this interchange. When everything is nonnegative (as it is here) that justification comes from the monotone convergence theorem and/or Tonelli's theorem.\r\n\r\n3. You still have to do some calculus, but it's fairly routine calculus.\r\n\r\n4. The statement is wrong. It should be $ \\sum_{n\\equal{}0}^{\\infty}\\frac1{(2\\plus{}3n)^2}.$", + "Solution_2": "Sorry , but are you sure about the index it should be from $ n \\equal{} 0$ ? \r\n\r\n\r\nIn fact, I could prove\r\n\\[ \\int\\limits_0^\\infty f(x) \\equal{} \\sum\\limits_{n\\equal{}1}^\\infty {\\int\\limits_{n \\minus{} 1}^{n} f(x)}\\]\r\nBut I stopped in calculus calculation ....Am I in the right way ?\r\n\r\nThank you :)", + "Solution_3": "[quote=\"Hidden Scofield\"]Sorry , but are you sure about the index it should be from $ n \\equal{} 0$ ? \n\n\nIn fact, I could prove\n\\[ \\int\\limits_0^\\infty f(x) \\equal{} \\sum\\limits_{n \\equal{} 1}^\\infty {\\int\\limits_{n \\minus{} 1}^{n} f(x)}\\]\nBut I stopped in calculus calculation ....Am I in the right way ?\n\nThank you :)[/quote]\r\n\r\nNo, you are in the wrong way. Do you know the corollary of the monotone convergence theorem that states $ \\int \\sum f_n \\equal{} \\sum \\int f_n$ whenever the $ f_n$ are nonnegative (and measurable)?", + "Solution_4": "Calculus: for $ a > 0,\\ \\int_0^{\\infty}xe^{ \\minus{} ax}\\,dx \\equal{} \\frac1{a^2}.$\r\n\r\nIt also looks like Hidden Schofield hasn't yet considered my point #1 above.", + "Solution_5": "[quote=\"Kent Merryfield\"]Calculus: for $ a > 0,\\ \\int_0^{\\infty}xe^{ \\minus{} ax}\\,dx \\equal{} \\frac1{a^2}.$\n\nIt also looks like Hidden Schofield hasn't yet considered my point #1 above.[/quote]\r\n\r\nIn fact , I don't know how to get Geometric series from that term .\r\n\r\nWWW:\r\n\r\nI know MCT and I made a mistake on my paper for $ 1\\plus{}e^{\\minus{}3x}$ instead of $ 1\\minus{}e^{\\minus{}3x}$.\r\n\r\nI still dizzy about what you need to point out for me ? Sorry .\r\n\r\nI don't need a direct answer , but I need clear way (more specific way) to do it by myself.\r\n :)", + "Solution_6": "The sum of an infinite geometric series is $ \\frac {\\text{first term}}{1 \\minus{} \\text{ratio}}$ provided the absolute value of the ratio is less than $ 1.$ Any time we see a fraction with a binomial denominator, we can try factoring out the larger piece of the denominator so we can write that denominator as $ 1$ minus something smaller - and we don't need to do that here, because the denominator is already in that form. Potential geometric series are everywhere, all the time - you just have to be aware of them.", + "Solution_7": "I got it ...Thx\r\n\r\nSo, $ f(x) \\equal{} \\sum\\limits_{n \\equal{} 0}^{\\infty}xe^{ \\minus{} (2 \\plus{} 3n)x}$, By MCT, we get the result .\r\n\r\nI'm so thankful ....thank you", + "Solution_8": "[quote=\"Hidden Scofield\"]\nWWW:\n\nI know MCT and I made a mistake on my paper for $ 1 \\plus{} e^{ \\minus{} 3x}$ instead of $ 1 \\minus{} e^{ \\minus{} 3x}$.[/quote]\r\n\r\nAs a further exercise, you should try the problem with $ 1 \\plus{} e^{\\minus{}3x}$ instead of $ 1 \\minus{} e^{ \\minus{} 3x}$.", + "Solution_9": "The MCT wouldn't apply directly for that version, but it's still fairly easy to justify interchanging sum and integral.", + "Solution_10": "We can take $ r\\equal{}\\minus{}e^{\\minus{}3x}$ when we have $ 1\\plus{}e^{\\minus{}3x}$\r\n\r\n[quote=\"Kent Merryfield\"]The MCT wouldn't apply directly for that version, but it's still fairly easy to justify interchanging sum and integral.[/quote]\r\n\r\nWhy ? the function inside the sum is measurable and nonnegative on $ (0,\\infty)$ for any $ n$, what are you trying to point for ?\r\n\r\nThank you all", + "Solution_11": "$ \\frac{xe^{\\minus{}2x}}{1\\plus{}e^{\\minus{}3x}}\\equal{}\\sum_{n\\equal{}0}^{\\infty}(\\minus{}1)^ne^{\\minus{}(2\\plus{}3n)x}$\r\n\r\nNot a sum of nonnegative terms.", + "Solution_12": "I got misunderstanding , I thought you talk about the first case of the function .\r\n\r\nIn second case, we can write the series as a difference of two series-es such that each one of them has positive term inside it. Then we take the integral to the difference, and we put the integral for each series and ...etc\r\n\r\nTRUE ?\r\n\r\nThank you :)", + "Solution_13": "Either that, or replace the terms by their absolute values, show that would give you an integrable sum (that's the first part of the problem, so we already know it works), and then cite the dominated convergence theorem.\r\n\r\nTheorem:\r\n\r\nSuppose $ \\sum_n\\int_S|f_n|<\\infty.$ Then $ \\int_S\\sum_nf_n \\equal{} \\sum_n\\int_Sf_n.$\r\n\r\nThe proof of this theorem involves both the DCT and the MCT. The dominating function for the DCT is $ g \\equal{} \\sum_n|f_n|.$ That $ g$ is integrable can be shown by the MCT.", + "Solution_14": "That's very nice .... :lol: \r\n\r\nThank you :)" +} +{ + "Tag": [], + "Problem": "[url]http://travel2.nytimes.com/2006/10/26/world/middleeast/26diplo.html[/url]\r\n\r\nThe draft of a U.N. security council resolution includes a provision that could potentially prohibit Iranian students abroad from studying physics at foreign universities. (Just how sweeping the prohibition would be depends on how you interpret it.) What do you think of this? Is it a great idea, or going too far? Better than economic sanctions? Better than your own ideas?\r\n\r\nPersonally, I'm a bit conflicted. Ordinarily, I would probably think it's a great idea that targets its impact as narrowly as possible. But when you have an Iranian friend and classmate who's planning on majoring in physics, you tend to shy away from supporting such measures!", + "Solution_1": "Uhm, would that also apply to doctorate student (phd students?), because I think that it would affect the mathematics department at my university then :!: \r\n\r\nI once met an Iranian student when I reached the last round of the physics olympiad. He had only been in Flanders for half a year, but he already spoke amazing Dutch. :( I really don't think he had bad intentions.", + "Solution_2": "[quote=\"fredbel6\"]Uhm, would that also apply to doctorate student (phd students?), because I think that it would affect the mathematics department at my university then :!: \n\nI once met an Iranian student when I reached the last round of the physics olympiad. He had only been in Flanders for half a year, but he already spoke amazing Dutch. :( I really don't think he had bad intentions.[/quote]\r\n\r\nDo you think that a language change someone's mind or intention?\r\nif yes thank i am in germany and i can't speak good German.\r\nDoes it mean that i have bad intentions ?" +} +{ + "Tag": [ + "inequalities", + "linear algebra", + "matrix", + "LaTeX" + ], + "Problem": "[b]Let[/b] $ A,B,C \\in M_{n} \\mathbb{(R)}$,$ I_{n}$[b] is identity nxn matrix .[/b]\r\n[b]Such that :[/b] $ A^{3} = (19 + 5).A^{2} - 19.5.A , 19.5.B^{3} = (19 + 5).\\pi .B^{2} - \\pi^{2}.B$\r\n[b]Prove that :[/b]\r\n$ det(e^{A} + x.I_{n}).det(e^{A} + y.I_{n}) \\geq det^{2}(e^{A} + \\sqrt {xy}.I_{n}),\\forall x,y \\in R^{ + }$\r\n$ det(Sin{B} + x.I_{n}).det(Sin{B} + y.I_{n}) \\geq det^{2}(Sin{B} + \\sqrt {xy}.I_{n}),\\forall x,y \\in R^{ + }$\r\n$ det(Ch(C)) \\geq 0$\r\n[b]\"... I have roused from Buddhism and love native place ,love country of Ho Chi Minh President ..\"[/b]", + "Solution_1": "dude, what the f*ck is this...\r\n\r\nfirst: if you're older than 18 years, you have some serious issues, act mature...\r\nsecond: this type of problems are really terrible. I can only guess you make those up yourself since no book ever would publish them\r\nthird: notice that almost nobody answers your question. Think why?", + "Solution_2": "Halmos, this guy may be no Ramanujan, but criticizing someone on grounds of what type of problems they are interested in is just stupid. What are you going to do next, tell him that his favorite band sucks? That he likes dumb food?\r\n\r\nIf you're frustrated, lodge a complaint with the moderators. There's a lot that can be done about the plague on our Linear Algebra forum without resorting to name-calling.", + "Solution_3": "I just find it hard to believe he is serious. \r\n\r\nI highly suspect him to be trolling. Also his \"love math, love nature,...\" remarks are inappropriate I think.\r\n\r\n[quote]\nIf you're frustrated, lodge a complaint with the moderators.\n[/quote]\r\n\r\nI have statistics exams these days, so yeah, I'm a bit grumpy ;)", + "Solution_4": "[quote=\"Halmos\"]I just find it hard to believe he is serious. \n\nI highly suspect him to be trolling. Also his \"love math, love nature,...\" remarks are inappropriate I think.\n[/quote]\r\n\r\nI thought he was a troll, until I thought about the kind of time investment this would require. Though it's still possible.", + "Solution_5": "[quote=\"Halmos\"]dude, what the f*ck is this...\n\nfirst: if you're older than 18 years, you have some serious issues, act mature...\nsecond: this type of problems are really terrible. I can only guess you make those up yourself since no book ever would publish them\nthird: notice that almost nobody answers your question. Think why?[/quote]\r\n[b]I am 24 years , I live in Viet Nam , I am a student at noi university of techlonogy , Steel-iron , I do learning english and math , english of me so bad because i will learn english good .... \" ..My Name is Dao Van Gioi ..\"[/b]", + "Solution_6": "[quote=\"studen of jean\"][b]Let[/b] $ A,B,C \\in M_{n} \\mathbb{(R)}$,$ I_{n}$[b] is identity nxn matrix .[/b]\n[b]Such that :[/b] $ A^{3} = (19 + 5).A^{2} - 19.5.A , 19.5.B^{3} = (19 + 5).\\pi .B^{2} - \\pi^{2}.B$\n[b]Prove that :[/b]\n$ det(e^{A} + x.I_{n}).det(e^{A} + y.I_{n}) \\geq det^{2}(e^{A} + \\sqrt {xy}.I_{n}),\\forall x,y \\in R^{ + }$\n$ det(Sin{B} + x.I_{n}).det(Sin{B} + y.I_{n}) \\geq det^{2}(Sin{B} + \\sqrt {xy}.I_{n}),\\forall x,y \\in R^{ + }$\n$ det(Ch(C)) \\geq 0$\n[b]\"... I have roused from Buddhism and love native place ,love country of Ho Chi Minh President ..\"[/b][/quote]\r\nSolution :\r\n Exitsts $ T \\in GL_{n} \\mathbb{(R)}$ such that :\r\n $ A =T\\[ diag\\] (19,19,19,......,19,5,5,5....,5,0,0,0,....,0)$$ T^_{-1}$ ,Have $ s_{1}: 19,s_{2}: 5,n-s_{1}-s_{2} : 0$\r\n$ det(e^{A}+xI_{n})=x^{n-s_{1}-s_{2}}(x+19)^{s_{1}}(x+5)^{s_{2}}$\r\nafter you can use inequality as :\r\n $ (x+a)(y+a) \\geq (\\sqrt {xy}+a)^{2}$, $ \\forall x,y,a \\geq 0$", + "Solution_7": "[quote=\"studen of jean\"][quote=\"studen of jean\"][b]Let[/b] $A,B,C \\in M_{n} \\mathbb{(R)}$,$I_{n}$[b] is identity nxn matrix .[/b]\r\n[b]Such that :[/b] $A^{3} = (19 + 5).A^{2} - 19.5.A , 19.5.B^{3} = (19 + 5).\\pi .B^{2} - \\pi^{2}.B$\r\n[b]Prove that :[/b]\r\n$det(e^{A} + x.I_{n}).det(e^{A} + y.I_{n}) \\geq det^{2}(e^{A} + \\sqrt {xy}.I_{n}),\\forall x,y \\in R^{ + }$\r\n$det(Sin{B} + x.I_{n}).det(Sin{B} + y.I_{n}) \\geq det^{2}(Sin{B} + \\sqrt {xy}.I_{n}),\\forall x,y \\in R^{ + }$\r\n$det(Ch(C)) \\geq 0$\r\n[b]\"... I have roused from Buddhism and love native place ,love country of Ho Chi Minh President ..\"[/b][/quote]\r\nSolution :\r\n Exitsts $T \\in GL_{n} \\mathbb{(R)}$ such that :\r\n $A = T$$diag$$(19,19,19,......,19,5,5,5....,5,0,0,0,....,0)$$T^_{ - 1}$ ,Have $s_{1}: 19,s_{2}: 5,n - s_{1} - s_{2} : 0$\r\n$det(e^{A} + xI_{n}) = x^{n - s_{1} - s_{2}}(x + 19)^{s_{1}}(x + 5)^{s_{2}}$\r\nafter you can use inequality as :\r\n $(x + a)(y + a) \\geq (\\sqrt {xy} + a)^{2}$, $\\forall x,y,a \\geq 0$[/quote]\r\n\r\nHuh? Are you sure that your LaTeX is correct? I couldn't understand your \"solution.\"\r\n\r\nPS: It says that \"You have BBCode (including possibly LaTeX) inside one of your formulas.\" so I don't know what to do but to disable BBCode in this post. How to deal with this? :huh:", + "Solution_8": "It's the double dollar signs. They indicate math mode on its own line, and don't mix with inline math mode.", + "Solution_9": "[quote=\"jmerry\"]It's the double dollar signs. They indicate math mode on its own line, and don't mix with inline math mode.[/quote]\r\n\r\nWhat is a \"line\" in this case (& what's \"inline)? I'm not a computer expert... :oops: \r\n\r\n\\[ LaTeX\\], is that it?" +} +{ + "Tag": [ + "inequalities proposed", + "inequalities" + ], + "Problem": "Prove that for positive real numbers $ a_1,\\dots,a_n$\r\n\\[ \\sum_{k\\equal{}1}^{n}\\sqrt[k]{a_{1}\\dots a_k}\\leq e\\sum_{k\\equal{}1}^{n}a_k\\]", + "Solution_1": "[quote=\"Yuriy Solovyov\"]Prove that for positive real numbers $ a_1,\\dots,a_n$\n\\[ \\sum_{k \\equal{} 1}^{n}\\sqrt [k]{a_{1}\\dots a_k}\\leq e\\sum_{k \\equal{} 1}^{n}a_k\\]\n[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=75674" +} +{ + "Tag": [ + "geometry", + "circumcircle", + "geometric transformation", + "homothety", + "geometry unsolved" + ], + "Problem": "PROBLEM(Lokman G\u00d6K\u00c7E): $D,E,F$ are mid-point on the arcs $BC,AC,AB$ of the circumcircle of $\\triangle ABC$.The excircle of $\\triangle ABC$ with respect to vertex $B$ is tangent to the sides at $K,L,M$.Reflection of $L$ is $L'$ witr respect to the excenter $I_{b}$\r\na) Show that $DK,FM,EL'$ are concurrent.\r\nb) If $DK\\bigcap FM=\\{X \\}$ and $O$ is the circumcenter then, show that $O,I_{b},X$ are collinear.", + "Solution_1": "Suppose that $FM$\u2229$OI_{b}$=${P}$,and the radius of \u2299$O$ and \u2299$I_{b}$ are $R$and$r$ respectively.\r\n\r\nBecause $F$ is the mid-point of arc $AB$,we can have $OF$\u22a5$BM$,and on another hand,we know that $I_{b}M$\u22a5$BM$ obviously.Hence,$OF$//$I_{b}M$,so $PI_{b}/PO$=$I_{b}M/OF$=$r/R$.\r\n\r\nIf we suppose $DK$\u2229$OI_{b}$=${P'}$and$EL'$\u2229$OI_{b}$=${P''}$,by imitating the process above,we can also have $P'I_{b}/P'O$=$r/R$and $P''I_{b}/P''O$=$r/R$.\r\n\r\nBecause of $PI_{b}/PO$=$P'I_{b}/P'O$= $P''I_{b}/P''O$=$r/R$,we can conclude that $P$,$P'$and$P''$ are the same point.\r\n\r\nThus,$a)$ and $b)$ are solved together.\r\n\r\nP.S:In fact,P is the center of similitude of \u2299$O$ and \u2299$I_{b}$ .", + "Solution_2": "Since $LL'$ is diameter of the excircle, $m(LML')=m(LKL')=90^{o}$.We see that $FD//MK,FD//ML',DE//KL'$. So, $\\triangle DEF$ and $\\triangle KL'M$ are homotethic.Therefore, $DK,EL',FM$ are concurrent.The intersection point of $DK,EL',FM$ is $X$.That is, $X$ is the homotety center.Then, circumcircles of $\\triangle DEF$ ,$\\triangle KL'M$ and the homotethy center are collinear!\r\n\r\nLokman G\u00d6K\u00c7E" +} +{ + "Tag": [ + "induction", + "algebra unsolved", + "algebra" + ], + "Problem": "May anyone help?This haunts me long.\r\n$\\displaystyle \\sum_{i=0}^n {C_n}^i=\\frac{n+1}{2^{n+1}}\\sum_{i=0}^n \\frac{2^{i+1}}{i+1}$\r\nHere,$\\displaystyle {C_n}^i=\\frac{n!}{i!(n-i)!}$", + "Solution_1": "We must prove, that:\r\n$\\sum_{k=0}^{n} \\frac{1}{\\binom{n}{k}}=\\frac{n+1}{2^{n+1}}\\cdot \\sum_{k=1}^{n+1} \\frac{2^k}{k}$\r\n\r\nLet \r\n$\\sum_{k=0}^{n} \\frac{1}{\\binom{n}{k}}=D_n$\r\n$\\sum_{k=0}^{n} \\frac{k+1}{\\binom{n}{k}}=E_n$\r\n\r\n$E_n=\\frac{1}{\\binom{n}{0}}+\\frac{2}{\\binom{n}{1}}+\\cdots +\\frac{n+1}{\\binom{n}{n}}$\r\n$E_n=\\frac{n+1}{\\binom{n}{n}}+\\frac{n}{\\binom{n}{n-1}}+\\cdots+\\frac{1}{\\binom{n}{0}}$\r\nWe add those equations and obtain:\r\n$2E_n=(n+2)D_n$\r\n\r\nNotice, that:\r\n$E_n=\\sum_{k=0}^{n} \\frac{k+1}{\\binom{n}{k}}=(n+1)\\cdot \\sum_{k=0}^{n} \\frac{1}{\\binom{n+1}{k+1}}=(n+1)(D_{n+1}-1)$\r\n\r\nBy combining with the previous equation we get:\r\n$D_{n+1}=1+\\frac{n+2}{2n+2}D_n$\r\n\r\nFrom here you can easily prove by induction, that \r\n$D_n=\\frac{n+1}{2^{n+1}}\\cdot \\sum_{k=1}^{n+1} \\frac{2^k}{k}$" +} +{ + "Tag": [], + "Problem": "In the equation shown, each letter represents a different digit. If A if not zero, what is the largest possible value of D.\r\n\r\n$ABC \\times C = DBC$.", + "Solution_1": "[quote=\"Ubemaya\"]In the equation shown, each letter represents a different digit. If A if not zero, what is the largest possible value of D.\n\n$ABC \\times C = DBC$.[/quote]\r\n\r\n[hide]\nC cannot be 1, because if it is 1, then D=A\nC is either 5 or 6\nif C is 5, then b iss either 2 or 7.\nif c is 6, then b can't be b\nwhen b is 2, it is $A25 \\times 5 = D25$. D could be 6 (1 more than a multiple of 5 since 25 x 5 = 125), then A is 1\nwhen b is 7, it is $A75 \\times 5 = D75$. D could be 8 (3 more than a multiple of 5 since 75x5=375), then A is 1\n\nthe highest possible value of $D$ is $8$\n[/hide]", + "Solution_2": "[hide]Answer = $8$\n\nYou can find that the only numbers that, when multiplied by themselves, have a ones digit of that number, are $1$, $5$, and $6$. It cannot be one, because any number times $1$ is itself, and we cannot have the same number be the product. So that leaves $5$ and $6$. \n\nWhen using $5$ as the $C$ number, you get a \"carryover\" of $2$ onto the next multiplication. That means that $D*5+2$ must have a units digit of $D$. The only numbers that this works for are $2$ and $5$. If you try the same thing with $6$, since $6*6$ has carryover of $3$, no numbers work ($D*6+3$ can never $=D$). \n\nThis leaves us with 18 possible choices. However, the product must also be less than 1000, because the result must be 3 digits. This leaves up with 2 possibilities, the larger of which, is $175*5=875$. The $D$ integer is $8$.\n\n[/hide]", + "Solution_3": "Both of you are correct!" +} +{ + "Tag": [ + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "Let $ a_n$ denote the number of nonempty sets $ S$ such that:\r\n(i)$ S \\subseteq \\{1,2,...,n\\}$\r\n(ii)all elements of $ S$ have the same parity;\r\n(iii)each element $ k\\in S$ such that $ k\\geq 2|S|$.\r\nProve that: $ a_{2m \\minus{} 1} \\equal{} 2(F_{m \\plus{} 1} \\minus{} 1)$ and $ a_{2m} \\equal{} F_{m \\plus{} 3} \\minus{} 2$\r\nWhere for all $ m\\geq 1$,$ F_m$ is the $ m^{th}$ of Fibonacci number ($ F_1 \\equal{} F_2 \\equal{} 1,F_{n \\plus{} 2} \\equal{} F_{n \\plus{} 1} \\plus{} F_n$ )\r\n\r\nI had using Properties of Binomial Coefficents to problem formula $ a_{n \\plus{} 4} \\equal{} a_n \\plus{} a_{n \\plus{} 2} \\plus{} 2$ but I think that them very complex.So I am wating a nice solution.", + "Solution_1": "[quote=\"math10\"]Let $ a_n$ denote the number of nonempty sets $ S$ such that:\n(i)$ S \\subseteq \\{1,2,...,n\\}$\n(ii)all elements of $ S$ have the same parity;\n(iii)each element $ k\\in S$ such that $ k\\geq 2|S|$.\nProve that: $ a_{2m \\minus{} 1} \\equal{} 2(F_{m \\plus{} 1} \\minus{} 1)$ and $ a_{2m} \\equal{} F_{m \\plus{} 3} \\minus{} 2$\nWhere for all $ m\\geq 1$,$ F_m$ is the $ m^{th}$ of Fibonacci number ($ F_1 \\equal{} F_2 \\equal{} 1,F_{n \\plus{} 2} \\equal{} F_{n \\plus{} 1} \\plus{} F_n$ )\n\nI had using Properties of Binomial Coefficents to problem formula $ a_{n \\plus{} 4} \\equal{} a_n \\plus{} a_{n \\plus{} 2} \\plus{} 2$ but I think that them very complex.So I am wating a nice solution.[/quote]\nSolution?", + "Solution_2": "[b]My Solution:[/b]\n[hide]\nLet ${T_n = \\{S \\in \\{1,2,\\cdots,n\\} \\ ; \\ S \\ \\mathrm{satisfies}\\ \\mathrm{ii)} \\ \\mathrm{and} \\ \\mathrm{iii)}} \\}$.\nWe partition $T_{n+4}$ into $3$ parts.\n1) $A_{n+4} = \\{S \\in T_{n+4} \\ ; \\ 1,2 \\notin S, \\ \\forall k \\in S, k \\geq 2|S|+2\\}$\nConstruct $f \\ : \\ \\mathcal{P}(\\{1,2,\\cdots,n+2\\}) \\rightarrow \\mathcal{P}(\\{1,2,\\cdots,n+4\\})$ satisfies that,\n\\[ f(\\{x_1,x_2,\\cdots,x_k\\}) = \\{x_1+2,x_2+2,\\cdots,x_k+2\\}. \\]\nObviously it's bijective and $f(T_{n+2}) = A_{n+4}$. So $|A_{n+4}|=|T_{n+2}|$.\n2) $B_{n+4} = \\{S \\in T_{n+4} \\ ; \\ 1,2 \\notin S, \\ \\exists k \\in S, k < 2|S|+2\\}$\nConstruct $f \\ : \\ \\mathcal{P}(\\{1,2,\\cdots,n\\}) \\rightarrow \\mathcal{P}(\\{1,2,\\cdots,n+4\\})$ satisfies that,\n\\[{{ \nf(\\o) = \\{3\\} \\\\\nf(\\{x_1,x_2,\\cdots,x_k\\}) = \\{x_1+4,x_2+4,\\cdots,x_k+4,2k\\}\\ ,\\mathrm{for} \\ \\mathrm{even}} \\\\\nf(\\{x_1,x_2,\\cdots,x_k\\}) = \\{x_1+4,x_2+4,\\cdots,x_k+4,2k+1\\}\\ ,\\mathrm{for}\\ \\mathrm{odd}}.\n\\]\nSimilarly, we have $|B_{n+4}|=|T_n|+1$.\n3) $C_{n+4} = \\{ S \\in T_{n+4} \\ ; \\ 1 \\in S \\ \\mathrm{or} \\ 2 \\in S \\}$\n$|C_{n+4}| = 1$ is self-evident.\nSo we get $|T_{n+4}|=|T_{n+2}|+|T_n|+2$. Then it's simple to prove that $a_{2m-1}=2(F_{m+1}-1)$ and $a_{2m} = F_{m+3}-2$.\n[/hide]", + "Solution_3": "We consider the case that $n$ is even.\nIf $|S|=1$ then $k \\ge 2$. Hence, the elements in $S$ can be one of $n-1$ numbers $2,3, \\cdots , n$, which have $\\frac n2$ even numbers and $\\frac{n-2}{2}$ odd numbers. This follows that there are $\\binom{\\frac n2}{1}+ \\binom{ \\frac n2 -1}{1}$ ways to create set $S$ such that $|S|=1$.\nIf $|S|=2$ then $k \\ge 4$. Hence, the elements in $S$ can be two of $n-3$ numbers $4,5,6, \\cdots , n$, which have $\\frac n2 -1$ even numbers and $\\frac n2 -2$ odd numbers. This follows that there are $\\binom{\\frac n2 -1}{2}+ \\binom{\\frac n2 -2}{2}$ ways to create set $S$ such that $|S|=2$.\nIf $|S|=3$ then $k \\ge 6$. Similarly, we get $\\binom{\\frac n2 -2}{3}+ \\binom{\\frac n2 -3}{3}$ ways to create set $S$ such that $|S|=3$.\nIf $|S|=4$ then there are $\\binom{\\frac n2 -3}{4}+ \\binom{ \\frac n2 -4}{4}$ ways to create set $S$ such that $|S|=4$.\n.......\nIf $|S|= k$ then there are $\\binom{\\frac n2 -k+1}{k}+ \\binom{\\frac n2 -k}{k}$ ways to create set $S$ such that $|S|=k$.\nThus, $a_n= \\sum_{k=0}^{\\frac n2} \\binom{ \\frac n2 -k+1}{k}+ \\sum_{k=0}^{\\frac n2} \\binom{\\frac n2 -k}{k}$ with $2|n$.\nUsing induction, we can see that $\\sum_{k=0}^{\\frac n2} \\binom{ \\frac n2 -k+1}{k}= F_{\\frac n2 +1}$ and $ \\sum_{k=0}^{\\frac n2} \\binom{\\frac n2 -k}{k}=F_{ \\frac n2 +2}-2$. Hence, $a_n=F_{\\frac n2+1}+F_{\\frac n2 +2}-2=F_{ \\frac n2 +3}-2$ or $a_{2m}=F_{m+3}-2$.\nSimilarly, we can prove that $a_{2m-1}=2(F_{m+1}-1)$.", + "Solution_4": "Actually, in your solution, [b]shinichiman[/b] , $k$ cannot be $0$ so the sum begins with $k=1$ and ends with $k=\\frac{n}{2}$ .\n\nTo prove the final statements, we should use the formula:\n\n$$F_n=\\sum_{k=0}^{[ \\frac{n-1}{2}]} \\binom{n-k-1}{k}$$\n\n, with $n=m+1$ and $n=m+2$.\n\nThen we must subtract $2$ to obtain the result." +} +{ + "Tag": [ + "number theory unsolved", + "number theory" + ], + "Problem": "- $ \\frac {2005}{2004^2\\plus{}1}$ +$ \\frac {2005}{2004^2\\plus{}2}$ +............+ $ \\frac {2005}{2004^2\\plus{}2004}$ isn't a natural number. :?:", + "Solution_1": "[quote=\"creatorvn\"]- $ \\frac {2005}{2004^2 \\plus{} 1}$ +$ \\frac {2005}{2004^2 \\plus{} 2}$ +............+ $ \\frac {2005}{2004^2 \\plus{} 2004}$ isn't a natural number. :?:[/quote]\r\n\r\n$ 2004 \\times \\frac{2005}{2004^2\\plus{}2004} \\equal{} 1 \\le \\frac{2005}{2004^2\\plus{}1}\\plus{}\\frac{2005}{2004^2\\plus{}2}\\plus{}\\cdots\\plus{}\\frac{2005}{2004^2\\plus{}2004} \\le 2004 \\times \\frac{2005}{2004^2\\plus{}1} \\le \\frac{2004\\times2005}{(2004\\plus{}1)(2004\\minus{}1)} \\equal{} \\frac{2004}{2003} \\leq 2$", + "Solution_2": "This fraction is $ \\frac {2005}{2004^2\\plus{}1}$ x -1" +} +{ + "Tag": [ + "inequalities", + "number theory proposed", + "number theory" + ], + "Problem": "Let $ a,b$ be positive integers such that $ a>b$ and $ n$ be a positive integer such that $ \\frac{b+n}{a}$ and $ \\frac{a+n}{b}$ are positive integers. if $ d=\\gcd(a,b)$ prove that:\r\n\r\n$ 2d\\leq (n+1)\\sqrt{a-b}$", + "Solution_1": "let $ a=da_{0}$ and $ b=db_{0}$. it is obvious that $ d|n$. if $ n\\geq 2d$ then our inequality is easily true so suppose $ n=d$. then we must have $ a_{0}|b_{0}+1$ abd $ b_{0}|a_{0}+1$.since $ a>b$ we have $ a_{0}+1>b_{0}+1\\geq a_{0}$ so $ b_{0}+1=a_{0}$ and we have that $ b_{0}|a_{0}+1=b_{0}+2$ so $ b_{0}=1$ and $ a_{0}=2$ or $ b_{0}=2$ and $ a_{0}=3$ now we need to check that $ 2d\\leq (d+1)\\sqrt{d}$which is true because $ d+1\\geq 2\\sqrt{d}$ by $ AM-GM$.\r\nequality cases: $ a=2,b=1$ and $ a=3,b=2$ where in both cases $ n=d=1$." +} +{ + "Tag": [ + "trigonometry", + "function", + "rotation" + ], + "Problem": "Hey guys please help me out on this. My book has the answers already, I just need to kno how to do it.\r\n\r\nF(x)= 3cos2 (x+pi/2) - 1 for x E {-pi/2, pi/2}\r\n\r\nF (x)= 2sin 2 (x+pi) + 1 for x E (-pi, pi)\r\n\r\nI found only one zero ( -3.40) for the first but the book has many zeros for it. It sais {-5pi/12, -pi/12, 7pi/12, 11pi/2} are the zeros.\r\n\r\nand for the cos function, I found -0.9552, but the book also has 0.9552 as a zero.\r\n\r\nCould anyone help me plz??", + "Solution_1": "[b]3cos2 (x+pi/2) - 1 = 0[/b]\r\nYou are looking for x such that\r\ncos2 (x+pi/2) = 1/3\r\nThe function cos2 (x+pi/2) has a period of pi, and will hit 1/3 twice in an interval of length pi, like the one you're given (once on the way up and again on the way down). It could do it 3 times if two of them were at the extremes of the interval, but that is not the case here.\r\nAlso\r\ncos2(x+pi/2)=cos(2x-pi)=cos(-2x-pi) is an even function. That means that if you find one solution, changing the sign will give you the other solution.\r\n\r\n[b]2sin2(x+pi) + 1 = 0[/b] means\r\nsin2(x+pi) = -1/2 or\r\nsin(2x+2pi) = -1/2\r\nSince the period of sin(2x+2pi) is pi, you should expect 4 solutions in the 2pi interval given.\r\nThat will happen when 2x+2pi = -30 degrees or\r\n2x+2pi = -pi/6 or 11pi/6 (and infinite more times at 2pi intervals).\r\nIt also happens for -(180-30) degrees = -150 degrees or \r\n-(pi-pi/6) = -5pi/6 or 7pi/6 (and infinite more times at 2pi intervals).\r\n\r\n2x+2pi = 11pi/6 when\r\n2x = 11pi/6 - 2pi = -pi/6 or x = -pi/12\r\nAdding pi we get another solution:\r\nx = -pi/12 + pi = 11pi/12\r\n\r\n2x+2pi = 7pi/6 when\r\n2x = 7pi/6 - 2pi = -5pi/6 or x = -5pi/12\r\nAdding pi we get another solution:\r\nx = -5pi/12 + pi = 7pi/12", + "Solution_2": "[quote=\"KMST\"][b]3cos2 (x+pi/2) - 1 = 0[/b]\nYou are looking for x such that\ncos2 (x+pi/2) = 1/3\nThe function cos2 (x+pi/2) has a period of pi, and will hit 1/3 twice in an interval of length pi, like the one you're given (once on the way up and again on the way down). It could do it 3 times if two of them were at the extremes of the interval, but that is not the case here.\nAlso\ncos2(x+pi/2)=cos(2x-pi)=cos(-2x-pi) is an even function. That means that if you find one solution, changing the sign will give you the other solution.[/quote]\r\n\r\nIsnt Cos2(x+pi/2)= cos (2x + pi) and not cos (2x-pi)? And how do you know if its an even function or not?\r\n\r\nBy the way is there any easier solution for the sine function?", + "Solution_3": "The problem is that cos2(x+pi/2) is NOT even associative with cos (2x + pi) as cos2(x+pi/2) implies (cos2)(x+pi/2)...", + "Solution_4": "I understood that what was meant was:\r\n$ g(x)\\equal{} cos[2(x\\plus{} \\pi /2)]\\equal{}cos(2x\\plus{} \\pi)$\r\n[quote=\"helpmeplz\"]Isnt Cos2(x+pi/2)= cos (2x + pi) and not cos (2x-pi)? And how do you know if its an even function or not?[/quote]\nI did not explain properly before. :blush: I'll try again.\n$ g(x)$ is an even function because it complies with the definition of even function:\n$ g(x)\\equal{}g(\\minus{}x)$\n It happens that $ g(x)\\equal{}cos(2x\\plus{} \\pi)\\equal{}cos(\\minus{}2x\\plus{} \\pi)\\equal{}g(\\minus{}x)$ because if you rotate the same $ 2x$ angle counterclockwise $ (\\plus{}2x)$ or clockwise $ (\\minus{}2x)$ from an angle of $ \\pi$, you get angles symmetrical with respect to the x-axis, which will have the same cosine.\n$ g(0)\\equal{}\\minus{}1$ and the function is symmetrical about $ x\\equal{}0$.\n(It is also symmetrical about $ x\\equal{}\\pi$ and for all the other maxima and minima, which was in my mind at the time, but is irrelevant to the problem at hand).\n\n[quote=\"helpmeplz\"]By the way is there any easier solution for the sine function?[/quote]I might have explained myself poorly, and made the path to the solution look more complicated than it was. The final solution is the same, though. The problem was actually made \"simpler\" by restricting the values for $ x$ to the interval $ ( \\minus{}\\pi, \\pi)$. Without such a restriction, that kind of problem has infinity of solutions, usually expressed as something like:\r\n$ x\\equal{} kT \\plus{} x_1$ and/or $ x\\equal{} kT \\plus{} x_2$ with $ T$ being the period of the function.\r\nFor example, $ sinx\\equal{}1/2$ happens for $ x\\equal{}\\pi /6$ in the first quadrant, and for $ x\\equal{}5\\pi /6$ in the second quadrant. If you add (or subtract) any multiple of $ 2\\pi$ (the period), you get \"co-terminal angles\" that have the same sine. If they do not give you limits for the $ x$'s, you end up having to state the infinity of solutions in a generalized way such as\r\n\"$ x\\equal{}\\pi /6 \\plus{} 2k\\pi$ [i][b]or[/b][/i] $ x\\equal{}5\\pi /6 \\plus{} 2k\\pi$, where $ k$ is any integer.\"" +} +{ + "Tag": [], + "Problem": "Convert Phenol to thio phenol...", + "Solution_1": "Use thallunation from suitable starting product :)", + "Solution_2": "I am asking a conversion... u cant use any siuitable starting material....", + "Solution_3": "[hide=\"Answer\"]1. $ \\ce{(EtO)_2POCl}$\n2. $ \\ce{KNH2}$, $ \\ce{K}\\minus{}\\ce{NH3}$\n3. $ \\ce{NaNO2} / \\ce{HCl}$\n4. $ \\ce{HS\\minus{}}$[/hide]", + "Solution_4": "[quote=\"valeriummaximum\"]I am asking a conversion... u cant use any siuitable starting material....[/quote]\r\n\r\nI meant convert phenol to benzene first and then use thallunation :) . \r\n\r\n@Carcul: Is there any effective method for conversion of phenol to benzene (without using Zn dust) :maybe:", + "Solution_5": "[hide=\"Yes there are\"]Method I: \n\n1. $ Ar\\minus{}OH \\plus{} TsCl \\longrightarrow Ar\\minus{}OTs$\n\n2. $ Ar\\minus{}OTs \\plus{} NH_2NH_2/Pd \\longrightarrow Ar\\minus{}H$\n\nMethod II:\n\n1. $ Ar\\minus{}OH \\plus{} (EtO)_2POCl \\longrightarrow Ar\\minus{}OPO(OEt)_2$\n\n2. $ Ar\\minus{}OPO(OEt)_2 \\plus{} Ti/THF \\longrightarrow Ar\\minus{}H$[/hide]\r\n\r\nBy the way, what is \"thallunation\"?", + "Solution_6": "Process of treating Benzene with thallium trifluoroacetate to give $ \\ce{C6H5\\minus{}Tl(OCOCF3)2}$ .", + "Solution_7": "1. That process is called \"thallation\", not \"thallunation\". The element Tl is thallium, not thallunium.\r\n\r\n2. The correct name of the reagent $ \\ce{Tl(CF3CO2)3}$ is thallium tris(trifluoroacetate), not \"thallium trifluoroacetate\", which probably don't even exist.\r\n\r\n3. Thallation is commonly the name given to the reaction between the reagent in 2. with a generic benzenoid aromatic compound, not only benzene.", + "Solution_8": "[quote=\"Carcul\"]1. That process is called \"thallation\", not \"thallunation\". The element Tl is thallium, not thallunium.[/quote]\nSorry that's how it was given in my book :) \n\n[quote=\"Carcul\"]2. The correct name of the reagent $ \\ce{Tl(CF3CO2)3}$ is thallium tris(trifluoroacetate), not \"thallium trifluoroacetate\", which probably don't even exist.[/quote]\nAgain it was given so in my book.\n\n[quote=\"Carcul\"]3. Thallation is commonly the name given to the reaction between the reagent in 2. with a generic benzenoid aromatic compound, not only benzene.[/quote]\r\nSorry my bad if i said that :)", + "Solution_9": "In that case your book is wrong. One must always have a critical eye regarding what is written in books." +} +{ + "Tag": [ + "Euler", + "geometry", + "arithmetic sequence", + "geometry unsolved" + ], + "Problem": "Show that in a non-equilateral triangle, the following are equivalent :\r\n(a) the angles of the triangle in arithmetic progression\r\n(b) the common tangent to the nine point circle and the incircle is parallel to the Euler line\r\n\r\nI have succeeded in showing that (a) implies (b) by brute force but there must be some better solution\r\nAnd i havent been able to show that (b) implies (a)", + "Solution_1": "it is [url=http://www.mathlinks.ro/viewtopic.php?t=152247]here again[/url]" +} +{ + "Tag": [ + "algebra", + "polynomial", + "calculus", + "derivative" + ], + "Problem": "Could someone help me with this please. If possible, a step by step solution so i know how to go about solving it, as i have no idea. Thanks.\r\n\r\nSuppose that the polynomial equation P(x) = 0 has a double root at x = alpha. Show that P'(alpha) = 0 where\r\nP'(x) is the derivative of P(x).", + "Solution_1": "$P(x)=(x-\\alpha)^2*Q(x)$ so $P'(x)=2(x-\\alpha)*Q(x)+(x-\\alpha)^2*Q'(x)$ so $P'(\\alpha)=0$", + "Solution_2": "you could also state that if $P(a)=P'(a)=0$ then $x=a$ is a root with multiplicity of at least 2\r\n[hide=\"here is how i proved it\"]\nwe have $P(a)=0$ and $P(x)=(x-a)Q(x)$\nthen from this we have\n$P'(x)=(1)Q(x)+(x-a)Q'(x)$ \nthe only way that $P'(a)=0$ is if $Q(x)=(x-a)Q'(x)$\nwhen we plug in $x=a$ we get $Q(a)=(a-a)Q'(a)$\nthus $x=a$ must also be a root of $Q(x)$\nand we have proven that $x=a$ must be a root of multiplicity of at least 2.\n[/hide]" +} +{ + "Tag": [ + "function", + "limit", + "algebra", + "polynomial", + "algebra proposed" + ], + "Problem": "$(i)$ $f(x)$ is a real-function st for all $a,b>0$ :\r\n$\\lim_{n \\to +\\infty}f(a+nb)=+\\infty$\r\nIs it neccessary that $\\lim_{x \\to +\\infty}f(x)=+\\infty$?\r\n\r\n$(ii)$* Same question but replace $a+nb$ by $f(n)$ is a polynomial of $n$?", + "Solution_1": "Any idea? :(", + "Solution_2": "For the first one,I think the answer is no.\r\nI have posted a similar one :P", + "Solution_3": "What's the solution? We need to contruct function $f$.", + "Solution_4": "this problem has been unsolved for long time\r\nany idea?", + "Solution_5": "[quote=\"hungkhtn\"]$(i)$ $f(x)$ is a real-function st for all $a,b>0$ :\n$\\lim_{n \\to+\\infty}f(a+nb)=+\\infty$\nIs it neccessary that $\\lim_{x \\to+\\infty}f(x)=+\\infty$?\n\n$(ii)$* Same question but replace $a+nb$ by $f(n)$ is a polynomial of $n$?[/quote]\r\n$(i)$\r\nThe answer would be \"no\" if the question was \"for a given pair $a,b>0$\", because one could construct a function that would have a different limit for every different pair $a,b>0$.\r\nBut since it's \"for all $a,b>0$\", then the answer is \"yes\": every $x$ can be expressed as a $a+nb$, so $f(x)$ can never be bounded (look at the definition of infinite limit).\r\n\r\n$(ii)$\r\nDoes the problem mean this?\r\n$f(x)$ is a real-function st for every $g(n)$ polynomial of $n$:\r\n$\\lim_{n \\to+\\infty}f(g(n))=+\\infty$\r\nIs it neccessary that $\\lim_{x \\to+\\infty}f(x)=+\\infty$?\r\nThen the answer is \"yes\" too. The argument is the same.", + "Solution_6": "Contrexample. Let $f(x)=x$ if $x\\not =p+q\\sqrt 2 , p,q \\in Z$ and $f(p+q\\sqrt 2 )=p^{2}-|q|$. If 2 terms $a+nb$ can be represent in form $p+q\\sqrt 2$, then $b=r_{1}+s_{1}\\sqrt 2, a=r_{2}+s_{2}\\sqrt 2, r_{i},s_{i}\\in Q$ and \\[lim_{n}f(a+nb)=\\infty\\] , but $f(n+n^{2}\\sqrt 2)=0.$ :D", + "Solution_7": "Which was the question?\r\n$\\lim_{n \\to+\\infty}f(a+nb)=+\\infty$ for all $a,b\\in\\mathbb{R}^{+}$ $\\Rightarrow$ $\\lim_{x \\to+\\infty}f(x)=+\\infty$ ?\r\nor\r\n\r\n$\\lim_{n \\to+\\infty}f(a+nb)=+\\infty$ for all $a,b\\in\\mathbb{R}^{+}$ $\\Leftarrow$ $\\lim_{x \\to+\\infty}f(x)=+\\infty$ ?", + "Solution_8": "I give example, when $\\forall a,b\\in R \\ \\lim_{n\\to \\infty}f(a+nb)=+\\infty$, but don't exist $\\lim_{x\\to \\infty }f(x)$.", + "Solution_9": "Sorry, I can't follow. You say $\\forall a,b\\in\\mathbb{R}$, but then you restrict $a,b\\in\\{\\mathbb{Q}+\\mathbb{Q}\\sqrt{2}\\}$.", + "Solution_10": "I say $f(x)=x$ if $x\\not =z+y\\sqrt 2 , \\ z,y\\in Z$, therefore $\\forall a,b\\in R$ exist $\\lim_{n\\to \\infty }f(a+nb)$." +} +{ + "Tag": [ + "combinatorial geometry", + "extremal principle" + ], + "Problem": "Let [i]M[/i] be a set of points on a plane such that every point in [i]M[/i] is the midpoint of two points in [i]M[/i]. Show that [i]M[/i] is an infinite set.", + "Solution_1": "[hide]If the set were finite, then it would have a convex hull[/hide]", + "Solution_2": "[hide=\"Equivalently\"] Pick a point in $ M$, and consider a point $ P$ of maximal distance from $ M$. [/hide]", + "Solution_3": "These types of problems are trivial if you have heard of the extremal principle.\r\n\r\n[u]Solution 1.[/u] Assume to the contrary that the set is finite. Id est, there exist two points $ A,B$ such that $ |AB|$ is the maximum distance between any two given points in the set. Then $ B$ is the midpoint of a segment. Suppose that the endpoints of the segment for which $ B$ is its midpoint has endpoints $ C,D$. Then one of $ \\triangle ABC$ and $ \\triangle ABD$ has an angle $ \\geq 90^{\\circ}$. Hence, either $ AC > AB$ or $ AD > AB$. Contradiction! Therefore, the set has infinite cardinality. $ \\Box$\r\n\r\n[u]Solution 2.[/u] Assume to the contrary that the set $ M$ is finite. Then $ M$ must have a convex hull $ \\mathcal{C}(M)$ (k81o7). But since there exists a point $ \\in\\mathcal{C}$ that is the midpoint of a segment created from points in $ M$, then you will have a point $ x\\in M\\subset \\mathcal{C}(M)$, but $ x\\notin \\mathcal{C}(M)$. Contradiction! $ \\Box$", + "Solution_4": "What is the $ \\text{Extremal Principle}$?", + "Solution_5": "Extremal principle effectively tells you to look at a certain pair which maximizes something. Usually, if there is a contradiction you'll get it somehow from that pair.", + "Solution_6": "[quote=\"boxedexe\"]\n[u]Solution 2.[/u] Assume to the contrary that the set $ M$ is finite. Then $ M$ must have a convex hull $ \\mathcal{C}(M)$ (k81o7). But since there exists a point $ \\in\\mathcal{C}$ that is the midpoint of a segment created from points in $ M$, then you will have a point $ x\\in M\\subset \\mathcal{C}(M)$, but $ x\\notin \\mathcal{C}(M)$. Contradiction! $ \\Box$[/quote]\r\nI can do it like this!" +} +{ + "Tag": [ + "analytic geometry", + "calculus", + "geometry", + "vector", + "parameterization", + "parametric equation" + ], + "Problem": "How do you find the distance from a point (x,y) to a plane containing points (a,b), (c,d), and (e, f). \r\n\r\nif that becomes to weird, assume that in a-f, two of the x's and two of the y's are 0, leaving only 1 x coordinate, and 1 y coordinate.", + "Solution_1": "the distance is 0, because the xy axis only contains one plane, and that lane contains all of the coordinates (x,y).\r\n\r\nIf you meant the (x,y,z) coordinates, that would be harder. one question on it was even in my dad's calculus book.", + "Solution_2": "[quote=\"1=2\"]the distance is 0, because the xy axis only contains one plane, and that lane contains all of the coordinates (x,y).\n\nIf you meant the (x,y,z) coordinates, that would be harder. one question on it was even in my dad's calculus book.[/quote]\r\nA calculus book? Ha. Think about that on a geometry test.", + "Solution_3": "So, as shown in the topic below this one on plane to point problems, there are many methods, some of them generalizations to n-dimensions, some of them, like the one I'm about to explain, only for $R^{3}$:\r\n\r\nThe shortest distance between the point and the plane is the distance between the point and the projection of that point on the plane, that being said, we can proceed as follows.\r\n\r\n-Define the plane (scalar equation)\r\n-Find a vector perpendicular to this plane (cross product)\r\n-Define a line using this new vector and the point in question (vector equation ---> parametric equation)\r\n-Find the point of intersection between the line and the plane\r\n-Find the distance between this new point and the point in question\r\n\r\n\r\nIf you went through this process in general you would derive a formula for it (in fact the formula torajirou posted in the previous topic may be exactly that) but this is personally the way I like to think about it." +} +{ + "Tag": [ + "AMC" + ], + "Problem": "I was just wondering, do you guys find my practice meets helpful?\r\nOr, do u enjoy my trilogy of practice meets?\r\nGive me comments!\r\nWhen actual competitions start, i'll post those too\r\n\r\n\r\nBTW, the next set of questions will come out VERY soon.....\r\nI already have them on hand, just too busy to type them up.....", + "Solution_1": "I like them very much, but they're a bit too hard for me...oh well.\r\n\r\nP.S. Is it only me or is the top of the forum a bit too crowded...", + "Solution_2": "I was wondering where your 4th practice meet was.", + "Solution_3": "It's up now. This nomenclature is getting more and more complicated each day...", + "Solution_4": "what nomenclature?", + "Solution_5": "Nomenclature is the way things are named.", + "Solution_6": "no\r\ni kno what nomenclature is named\r\ni wanted to kno what mathfanatic was talking about my nomenclature getting weirder and weirder", + "Solution_7": "I think it's refering to the \"A junior meet, which is actually harder than senior\" problem series...either that or he's talking about himself abstractly :P", + "Solution_8": "In case you haven't noticed:\r\nAmazing (or whatever) starts with A\r\nBiology starts with B\r\nCrazy starts with C\r\n...\r\nHominid starts with H\r\nIntergalactic (or whatever) starts with I\r\nJello starts with J\r\n\r\n\r\nWHOA THERE'S A PATTERN HOW COOL.", + "Solution_9": "LOL. I didn't notice that till now.", + "Solution_10": "OMG, so [b]that[/b]'s the hidden secret behind those seemingly meaningless titles...COOLIO! :lol: \r\n\r\n...the first and the last time I use the words \"OMG\" and \"COOLIO\"...oh wait, I just said it again haven't I... :lol:", + "Solution_11": "Wow, MysticTerminator is vvvveeerrryyyy observant. Yeah, I can't believe he figured it all out by himself. Wowee!\r\n\r\nAnd yes, that's what I meant when I said the nomenclature was getting more and more complicated. Maybe if we just said \"problem #XXX\" or something. That would work provided we didn't post new problems at the same time (which rarely happens).\r\n\r\nP.S. Tare, your AoPS tips are good, but they're kind of condescending...n00bs wanna learn that stuff on their own, you don't have to attach it to everything you post.", + "Solution_12": "[quote=\"mathfanatic\"]Wow, MysticTerminator is vvvveeerrryyyy observant. Yeah, I can't believe he figured it all out by himself. Wowee![/quote]\r\n\r\nyeah...it's so simple, but i didn't notice it until a week ago or so. it took me a while. (Hmmm...what does gastrocnemius have to do with jello...hmmm...)", + "Solution_13": "I noticed it eventually. But why did you choose THOSE words rather than other words that began with those letters?", + "Solution_14": "what practice meets?", + "Solution_15": "If you look at older posts, you should see several series of questions making reference to \"Senior Meets\" or \"Practice Meets.\" Those are what he's talking about.", + "Solution_16": "[quote=\"ComplexZeta\"]I noticed it eventually. But why did you choose THOSE words rather than other words that began with those letters?[/quote]\r\n\r\nWell, I couldn't call them \"Algebra Challenge\" if it weren't algebra-related. Anyway, it added to the suspense. :wink:" +} +{ + "Tag": [ + "geometry" + ], + "Problem": "Tennyson is building a circular corral for his pet dinosaur. To form the corral he pounds 20 posts equally spaced around the circumference. Each post contributes 6 inches to the circumference and the fencing between adjacent posts contributes 8 ft to the circumference. To the nearest whole number, what is the number of square feet in the area of the corral? Express your answer to the nearest integer.\r\n\r\n[hide]2300[/hide]\r\n\r\nThank you.", + "Solution_1": "There are 20 spaces and posts, so 20*8 + 20(1/2) = 170... Then you do 170/2pi which is 27.05634033 and you square that and multiply times pi, and you get 2299.788928, which rounds up to [b]2300 ft^2[/b]" +} +{ + "Tag": [], + "Problem": "It takes Casey forever to clean her room. Forever is equal to 4 hours. She is a VERY slow worker. It takes her sister Janet 1/2 hour to clean the same room. If they both clean together, how long will it take them to clean her room. (I havent worked this one out yet so might not get a nice integer answer).", + "Solution_1": "[hide]If Casey takes 4 hours to clean her room, she cleans at a rate of $\\frac14$ per hour. In 1 hour, her sister does the equivolent of 2 rooms. Therefore, every hour $\\frac94$ of the room is cleaned. To clean 1 room will therefore take $\\frac{1}{\\frac94}=\\frac49$ hours. Man, the sister is fast and Casey is SLOW.[/hide]", + "Solution_2": "[hide]\nCasey does $\\frac{1}{4}$ of the room for every hour, her sister does $2$ rooms for every hour, which means that they do $\\frac{9}{4}$ of the room in an hour.\n$\\frac{1}{\\frac{9}{4}}=\\frac{4}{9}$ So, it takes them $\\frac{4}{9}$ of an hour to clean the room.[/hide]" +} +{ + "Tag": [ + "probability", + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "Hi Everybody ! \r\n\r\nExcuse my english please, I hope you will understand ! :roll: \r\n\r\nThere are 3 boxes and only one of them contains a surprise, the others are empty. 2 players are in this game, the first knows what is in every box and the second doesn't know anything. On the first move, the second must choose one of the boxes and take it. Then the first player who knows the containing of every boxes throw away an empty box among those which remained. After that the second player have to choose and take one of 2 boxes who remains. What is the best tactic for the second player for having the most chances to choose the surprise and what probability has it in that case ?\r\n\r\n-------------------------------------------French Party----------------------------------------------------------\r\nEt une petite traduction pour les francophones et les autres si c'est plus facile que mon anglais brouill\u00e9 ;) \r\n\r\nIl y a trois bo\u00eftes sur une table. Une d'entre elle contient une surprise (un bonbon :lol: ) et les autres sont vides. Il y a deux personnes, la premi\u00e8re qui conna\u00eft le contenu de chaque bo\u00efte et la deuxi\u00e8me qui ne sait rien. La deuxi\u00e8me personne choisit une boites. Ensuite la premi\u00e8re enl\u00e8ve du jeu une boite vide parmi les 2 qui restent. Maintenant la deuxi\u00e8me personne peut choisir entre une des 2 boites restantes. Quel est la meilleure technique que doit adopter le deuxi\u00e8me joueur pour maximiser ses chances de gagner la surprise et quelle serait sa probabilit\u00e9 dans ce cas ?\r\n\r\nJ'ai un mini doute en ayant \u00e9cout\u00e9 la solution d'un copain alors je fais apelle \u00e0 vous :bruce:\r\n\r\n[hide]My solution is 2/3 chances in this way : You choose first at random and then you only change your box... and you find 2/3..[/hide]", + "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?highlight=monty+hall&t=41658\r\n\r\nThis is both a famous and interesting problem.", + "Solution_2": "Thank you for link :thumbup:" +} +{ + "Tag": [ + "counting", + "distinguishability" + ], + "Problem": "If you want to assign 5 different jobs to 4 people, in how many ways can you do that? (Each person must get atleast 1 job)", + "Solution_1": ":blush: I stink at counting.", + "Solution_2": "But, doesn't the problem say that each person has to have [b]at least[/b] one job? So someone could have 2 jobs. :ninja:", + "Solution_3": "Are the jobs distinguishable from one another? If they are would the answser be 5!*5=600?", + "Solution_4": "[hide]I'm pretty sure that it is 240. One person must have two jobs. So...\n\n5C2 ways to pick 2 jobs x 4 people x 3! ways to assign the remaining jobs = 240\n[/hide]", + "Solution_5": "Hmm..I think my solutions wrong..number one.. theres 4 people not 5, so its 5!*4=480, but I doubled the job orderings for the person with two jobs (so AB was different from BA), so 480/2=240..wow i suck at math..", + "Solution_6": "nevermind.", + "Solution_7": "[hide]Give one job automatically to the four people. THere are 24 ways to arrange this for each set * 5 for the 5 combinations=120.\nThen there are 5 ways to assign the last digit, so we get 600.[/hide]", + "Solution_8": "[quote=\"bpms\"][hide]Give one job automatically to the four people. THere are 24 ways to arrange this for each set * 5 for the 5 combinations=120.\nThen there are 5 ways to assign the last digit, so we get 600.[/hide][/quote]\n[hide]The way to approach problems like this is to treat two jobs as a super job, and then assign the resulting 4 jobs to the 4 people. There are $\\binom{5}{2}$ to ways make a super job out of 5 jobs. The resulting 4 jobs can be assigned to 4 people in $4!$ ways. So the answer is $\\binom{5}{2}4! = \\boxed{240}$[/hide].", + "Solution_9": "That is the correct answer.\r\n\r\nGood Job", + "Solution_10": "wouldn't it be $5C2\\times3!$?", + "Solution_11": "[quote=\"sonny\"]If you want to assign 5 different jobs to 4 people, in how many ways can you do that? (Each person must get atleast 1 job)[/quote]\r\n\r\n[hide]The jobs are indistinguishable.\n\nSo one person must get 2 jobs.\n\nThere are 4 ways for that to happen\n\nSo the answer is\n\n$4*\\binom{5}{2}*3*2*1=240$[/hide]", + "Solution_12": "Ah, i get it! Anyway, I need help with counting, it gives me a headache." +} +{ + "Tag": [ + "calculus", + "derivative", + "induction", + "calculus computations" + ], + "Problem": "find the thousandth derivative of\r\n\r\nf(x) = x e^(-x)", + "Solution_1": "Hello\r\n[hide=\"solution\"]\nf(x) = x(e)^( - x) \n f'(x) = (1 - x)e^( - x)\n f''(x) =(x - 2) e^( - x)\nf'''(x) =(3 - x) e^( - x)\n\nThus thousandth derivative=(x - 1000)e^( - x)\n[/hide]\r\nThank u.", + "Solution_2": "I got it this way:\r\n\r\ny = x e^(-x)\r\n\r\ny' = -x e^(-x) + e^(-x)\r\n\r\n = - y + e^(-x)\r\n\r\ny'' = - y' - e^(-x)\r\n\r\ny''' = - y'' + e^(-x).\r\n\r\n...............................\r\n\r\ny(1000) = - y(999) - e^(-x).\r\n\r\nso,\r\n\r\ny + y' = e^(-x)\r\n\r\ny' + y'' = -e^(-x)\r\n\r\ny'' + y''' = e^(-x)\r\n\r\n........................\r\n\r\ny(1000) + y(999) = -e^(-x).\r\n\r\nsubtracting successive equations, y' , y'', y''', .... , y(999) cancel out giving,\r\n\r\ny(1000) + y = (1000) e^(-x)\r\n\r\ny(1000) = (1000)e^(-x) - xe^(-x)\r\n\r\n = e^(-x) (1000 - x). :lol:", + "Solution_3": "hello\r\nSorry but I think the answer is e^(-x)(x-1000) instead of e^(-x) (1000 - x).\r\nKindly correct me if I m wrong.\r\nThank u.", + "Solution_4": "Proof by induction:\r\n\r\n$ \\forall n > 1$ we have the following proposition: $ \\frac {d^{n}}{dx^{n}}x e^{ \\minus{} x} \\equal{} ( \\minus{} 1)^{n}(x \\minus{} n)e^{ \\minus{} x}$.\r\n\r\nFor $ n \\equal{} 1$ we have: $ \\frac {d^{1}}{dx^{1}}x e^{ \\minus{} x} \\equal{} \\frac {d}{dx}x e^{ \\minus{} x} \\equal{} e^{ \\minus{} x} \\minus{} x e^{ \\minus{} x} \\equal{} ( \\minus{} 1)^{1}(x \\minus{} 1)e^{ \\minus{} x}$, hence the proposition is true for $ n \\equal{} 1$.\r\n\r\nSuppose the proposition is true for $ n \\equal{} k$ for some integer $ k \\geq 1$. Then we get for $ n \\equal{} k \\plus{} 1$:\r\n\r\n$ \\frac {d^{n}}{dx^{n}}x e^{ \\minus{} x} \\equal{} \\frac {d^{k \\plus{} 1}}{dx^{k \\plus{} 1}}x e^{ \\minus{} x} \\equal{} \\frac {d}{dx}\\left[\\frac {d^{k}}{dx^{k}}x e^{ \\minus{} x}\\right] \\equal{} \\frac {d}{dx}\\left [ ( \\minus{} 1)^{k}(x \\minus{} k)e^{ \\minus{} x} \\right] \\equal{} ( \\minus{} 1)^{k} \\left [ \\left( \\frac {d}{dx} x e^{ \\minus{} x}\\right ) \\minus{} k \\left ( \\frac {d}{dx} e^{ \\minus{} x} \\right ) \\right ] \\equal{} ( \\minus{} 1)^{k} \\left [ \\left( e^{ \\minus{} x} \\minus{} x e^{ \\minus{} x} \\right ) \\minus{} k \\left ( \\minus{} e^{ \\minus{} x} \\right ) \\right ] \\equal{} ( \\minus{} 1)^{k} \\left [ (k \\plus{} 1)e^{ \\minus{} x} \\minus{} x e^{ \\minus{} x} \\right ] \\equal{} ( \\minus{} 1)^{k \\plus{} 1}\\left [x \\minus{} (k \\plus{} 1) \\right ] e^{ \\minus{} x}$.\r\n\r\nThus the proposition is also true for $ n \\equal{} k \\plus{} 1$. By induction we get that the proposition is true $ \\forall n > 1$. Done.\r\n(yeah it's also true for $ n \\leq 0$ but I thought $ \\frac{d^{0}}{dx^{0}}$ etc. was kind of ugly. Also it's unnecessary here.)\r\n\r\n\r\nSo we get $ \\frac {d^{1000}}{dx^{1000}}x e^{ \\minus{} x} \\equal{} ( \\minus{} 1)^{1000}(x \\minus{} 1000)e^{ \\minus{} x} \\equal{} (x \\minus{} 1000)e^{ \\minus{} x}$, as $ 1000$ is even ;).", + "Solution_5": "hello, i also have $ f^{(1000)}(x) \\equal{} \\exp( \\minus{} x)(x \\minus{} 1000)$.\r\nSonnhard.", + "Solution_6": "[quote=\"kabi\"]hello\nSorry but I think the answer is e^(-x)(x-1000) instead of e^(-x) (1000 - x).\nKindly correct me if I m wrong.\nThank u.[/quote]\r\n\r\nyes, you are right, the answer is e^(-x) (x - 1000). I did a little mistake." +} +{ + "Tag": [], + "Problem": "Fie numerele reale x,y,z.Demonstrati ca (x^5-x^2)/(x^5+y^2+z^2)+(y^5-y^2)/(y^5+z^2+x^2)+(z^5-z^2)/(z^5+x^2+y^2)>=0.", + "Solution_1": "cred ca ineg e gresita!!Daca iei spre exemplu x,y,z=0,2 nu se verififca(si exemple sunt multe)", + "Solution_2": "Mie mi-a dat-o un coleg la scoala s-o rezolv.E adevarat ce zici tu, pentru x,y,z=0,2 nu se verifica.Inseamna ca eu o demonstrasem gresit.Sa vad ce pot sa fac.Poate reusesc sa o imbunatatesc...", + "Solution_3": "Ma indoiesc... vedeti ca e problema de la IMO de anul trecut... gresita desigur caci lipseste ipoteza ca produsul numerelor e 1 (mai sunt 1000000 ipoteze posibile care ar face problema mai interesanta, dar oricum e prea urata ca sa-si mai bata cineva capul cu ea).", + "Solution_4": "Asa e Harazi!Am uitat sa spun ca xyz=1 :blush: :blush: Acum sigur merge..." +} +{ + "Tag": [ + "probability" + ], + "Problem": "What is the probability that a multiple of 2 is also a multiple of 3? Express your answer as a common fraction.", + "Solution_1": "1/3,\r\nsince 2 and 3 are relatively prime.", + "Solution_2": "That doesn't make a lot of sense.\r\nA multiple of 2 and 3 must be a multiple of 6. Clearly this is 1/6 of the numbers.\r\n\r\nEDIT: Oops I didn't realize it was just multiples of 2, not all numbers. :blush:", + "Solution_3": "dude.\r\nit's 1/3\r\n1/6 of all numbers,\r\nbut 1/3 of half of the numbers.\r\nit asked probability that a multiple of 2 is multiple of 3\r\nhere:\r\nprobability that $ 2n$ is a multiple of $ 3$\r\nyou can clearly see that the 2 can be ignored so it's\r\nprobability that $ n$ is a multiple of $ 3$, which is $ \\frac13$", + "Solution_4": "1/3 because every third multiple of is divisiblke by 6" +} +{ + "Tag": [], + "Problem": "For Eine Kleine Nachtmusik, how do you play the trills? I mean more specifically which notes do you play? Say for the trill on the first page with the \r\nA A A A B A. (hopefully you guys know what I'm talking about :rotfl: )", + "Solution_1": "http://www.oldflutes.com/articles/classicaltrill/index.htm", + "Solution_2": "Thanks, I know how to play it now :)" +} +{ + "Tag": [ + "Ring Theory", + "algebra", + "polynomial", + "number theory", + "greatest common divisor", + "Galois Theory", + "superior algebra" + ], + "Problem": "I'm working off a book, Galois Theory, by David A. Cox, and I'm not entirely sure if this is valid:\r\n\r\nProve that = {xg + yh : g, h are in F[x, y]} contained in F[x, y] is not a principal ideal domain.\r\n\r\nAssume that (xg + yh) generates for some polynomials g, h in F[x, y]. Then there must exist a polynomial f such that (xg + yh)f = x, since x is in . So it must be the case that h = 0, otherwise you would have a y term. The same reasoning stands that g = 0 since we need to generate y. But if g and h are 0, then you generate nothing but 0. Contradiction.", + "Solution_1": "No it's not valid. The conclusion $h=0$ is wrong. Remember that $g,h$ are polynomials in $x$ and $y$.\r\n\r\nBesides, the claim is probably that $< x , y >$ is no principial ideal. As a consequence, $F[x,y]$ is no principial ideal domain.\r\n\r\nThe correct proof: The ideal $$ consists of the polynomials without constant term. Especially, it's not $F[x,y]$. So, if $ = $ for some $g \\in F[x,y]$, then $g$ is no unit. On the other hand, it's a greatest common divisor of $x,y$. But then $g = ax$ and $g = by$ for some $a,b \\in F^{*}$, a contradiction.", + "Solution_2": "Ah, sorry, you're right. I copied the question wrong. Show that is not a principal ideal.\r\n\r\nBut I don't understand your reasoning for why I am wrong. I'm going to change up some of the variables I used in the first, as I did not notice they conflicted. \r\n\r\nIf we assume there is an element of that generates , then let that element be (xn + ym) for some n, m in F[x, y]. Then doesn't it have to be the case that there exists f in F[x, y] such that (xn + ym)f = x, because x is in ? Multiplying these out, we get xnf + ymf. \r\n\r\nThere is no way to get ride of the y term in ymf unless we make m or f equal 0, and since we can't make f = 0, it must be that m = 0. No?\r\n\r\nWhich part of that is wrong?", + "Solution_3": "[quote=\"Ricky\"]\nThere is no way to get ride of the y term in ymf unless we make m or f equal 0, and since we can't make f = 0, it must be that m = 0[/quote]\r\n\r\nThis is wrong.", + "Solution_4": "Ok, that was the part I was unsure about. But why is it wrong? Can we say that f = y^-1 or some variant of it?", + "Solution_5": "[quote]On the other hand, it's a greatest common divisor of x,y. But then g = ax and g = by for some $a,b \\in F^{*}$, a contradiction.[/quote]\r\n\r\nWouldn't that be x = ga and y = gb?\r\n\r\nI don't see how you can reach the conclusion that g must have an inverse from being the gcd of x and y." +} +{ + "Tag": [ + "calculus", + "integration", + "trigonometry", + "calculus computations" + ], + "Problem": "it looks to me that it is an inverse trig. I tried to use the u-sub but it was too complicated when i plug it in \r\n[img]http://img34.imageshack.us/img34/336/69945314.jpg[/img]", + "Solution_1": "[quote=\"Aafour\"]I tried to use the u-sub but it was too complicated when i plug it in [/quote]\r\n[i]What[/i] substitution, exactly, did you try?\r\n\r\nI always suggest to my class that the first place you look (before you even consider trig substitutions) is the \"natural substitution\" or to substitute for the \"thing inside\". In this case, that would be $ u\\equal{}25\\minus{}x^2.$ Did you try that, and if you did, what happened?", + "Solution_2": "Set $ 25 \\minus{} x^2 \\equal{} z^2$. Then $ \\minus{}zdz \\equal{} xdx$", + "Solution_3": "akech - while that does work, in my opinion it's seriously confusing to a beginner. It's not what I'd like to see Aafour try.", + "Solution_4": "And I do agree with you. The trig substitution $ x\\equal{}5\\sin \\theta$ can also be used. But sticking with the first substitution you suggested may be the best thing as that is in line with the \"u\" substitution most students of integral calculus are familiar with.", + "Solution_5": "thank you guys, i used 25-x^2 as Kent suggested and it did work." +} +{ + "Tag": [ + "probability" + ], + "Problem": "Two unit squares are randomly placed on a 10x10 unit grid. What is the probability of the two squares having at least one unoccupied square between them (including diagonally)?", + "Solution_1": "[hide=\"hint\"]how many ways can they be placed so there are NO spaces inbetween them.[/hide]", + "Solution_2": "[hide=\"I will give it a try...\"]\nUsing abacadaea's hint, first find the probability that there are no spaces in between the two squares and subtract that from 1.\n\n[b]Corner Squares[/b]\nThe probability of the first square being on the corner is $ \\frac{4}{100}$, or $ \\frac{1}{25}$.\nThe probability of the second square being adjacent (here including the diagonal case) is $ \\frac{3}{99}$, or $ \\frac{1}{33}$.\nMultiply these two probabilities together and you get $ \\frac{1}{825}$.\n\n[b]Edge Squares[/b]\nP(Edge) = $ \\frac{32}{100}$, or $ \\frac{8}{25}$.\nP(Adjacent to Edge) = $ \\frac{5}{99}$.\n$ \\frac{5}{99}*\\frac{8}{25}\\equal{}\\frac{8}{495}$.\n\n[b]Middle Squares[/b]\nP(Middle) = $ \\frac{64}{100}$, or $ \\frac{16}{25}$.\nP(Adjacent to Middle) = $ \\frac{8}{99}$.\n$ \\frac{16}{25}*\\frac{8}{99}\\equal{}\\frac{128}{2475}$.\n\nAdd these up: $ \\frac{1}{825}\\plus{}\\frac{8}{495}\\plus{}\\frac{128}{2475}\\equal{}\\frac{155}{2475}\\equal{}\\frac{31}{495}$.\n\nThen, subtract this from 1: $ 1\\minus{}\\frac{31}{495}\\equal{}\\frac{464}{495}$.\n\nAnd my answer is: $ \\boxed{\\frac{464}{495}}$[/hide]\r\n\r\nI'm tired... :sleeping:", + "Solution_3": "@ Kevin - The way you did it is correct, but simplifying all the fractions in the beginning actually caused more calculations to be done later.\r\n\r\n[hide=\"I would do...\"]$ \\frac{4}{100}\\cdot\\frac{3}{99}\\plus{}\\frac{32}{100}\\cdot\\frac{5}{99}\\plus{}\\frac{64}{100}\\cdot\\frac{8}{99}\\equal{}\\frac{654}{9900}$ and then simplify and finish off.[/hide]", + "Solution_4": "[hide=\"Now we are almost there\"] \n If we do that then we are not accounting for the \"one space apart\" on the edges and corners. Any square adjacent to a corner can only have three possible other squares that satisfy the requirements. Also, any square adjacent to an edge (besides a corner and the squares adjacent to a corner) has five squares that are one apart from it. we count the number of squares adjacent to corners, squares adjecent to edges and squares in the middle to get: \n\n$ ({16\\over 100}*{3\\over 99})\\plus{}({48\\over 100}*{5\\over 99})\\plus{}({36\\over 100}*{8\\over 99}) \\equal{}{48\\over 9900}\\plus{}{240\\over 9900}\\plus{}{288\\over 9900}\\equal{}{576\\over 9900}$\nNow we're golden. Simplify to\n$ {16\\over 275}$\n\n[/hide]", + "Solution_5": "@ dracs taxi - I don't clearly understand your solution, and your answer is incorrect. Kevin K.'s solution is correct.\r\n\r\nWhere did you get $ \\frac{16}{100}$, $ \\frac{48}{100}$, and $ \\frac{36}{100}$ from?\r\n\r\n[quote=\"dracs taxi\"]If we do that then we are not accounting for the \"one space apart\" on the edges and corners.[/quote]\r\nYes we are, because we are finding the number of placements of squares that don't work, and then taking the complement of that.", + "Solution_6": "Oh gosh I'm really sorry.\r\n I compleatly missed the, ' \"at least\" one square apart.' Totally my bad. Now I agree with you entirely. :blush:", + "Solution_7": "No need to be sorry, everbody makes mistakes sometimes and without them we wouldn't learn. :)" +} +{ + "Tag": [ + "inequalities", + "algebra", + "polynomial", + "calculus", + "derivative", + "function", + "inequalities unsolved" + ], + "Problem": "Let $ a,b,c\\geq0$ s.t. $ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$.Prove that:\r\n$ \\dfrac{1}{3\\plus{}a^2\\minus{}2bc}\\plus{}\\dfrac{1}{3\\plus{}b^2\\minus{}2ca}\\plus{}\\dfrac{1}{3\\plus{}c^2\\minus{}2ab}\\leq\\dfrac{9}{8}$", + "Solution_1": "Similar ineq (easier and classical) \r\n$ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$\r\n<=>$ \\frac{1}{4\\plus{}a^2\\minus{}2bc} \\le \\frac{9}{11}$\r\nSetting :$ x\\equal{}ab\\plus{}bc\\plus{}ca \\le 1$\r\n<=>$ \\sum \\frac{1\\plus{}a^2\\minus{}2bc}{4\\plus{}a^2\\minus{}2bc} \\ge \\frac{6}{11}$\r\nBy cauchy-schwarz , we can prove :\r\n$ LHS \\ge \\frac{6(2\\minus{}x)^2}{27\\minus{}15x\\minus{}3x^2\\plus{}2}$\r\nSo , we prove :\r\n$ \\frac{(2\\minus{}x)^2}{27\\minus{}15x\\minus{}3x^2\\plus{}2} \\ge \\frac{1}{11}$\r\n<=>$ (1\\minus{}x)(15\\minus{}14x) \\ge 0$(Right $ x \\le 1$)\r\n^^.\r\nMy solution for ineq master's ineq is similar the above solution but it have a little SOS tech ^^\r\n______________\r\nSorry for my English ;)", + "Solution_2": "Is it enough this to prove the initial inequality?\r\n :?: \r\nThanks.", + "Solution_3": ":D .It's not enough to prove the first ineq ^^ .But if we use Schur .It's so easy :)\r\n[b]Solution of chien_than[/b]\r\n\r\n<=>$ \\sum_{cyc} \\frac {1 \\plus{} a^2 \\minus{} 2bc}{3 \\plus{} a^2 \\minus{} 2bc} \\geq \\frac {3}{4}$\r\nUse cauchy-schwarz , we have :\r\n$ \\sum_{cyc} \\frac {1 \\plus{} a^2 \\minus{} 2bc}{3 \\plus{} a^2 \\minus{} 2bc} \\geq \\frac {4(2 \\minus{} \\sum bc)^2}{8 \\minus{} 4\\sum bc \\plus{} \\sum(1 \\plus{} a^2 \\minus{} 2bc)^2}$\r\nWe need to prove\r\n$ 16(2 \\minus{} \\sum bc)^2\\geq 24 \\minus{} 12\\sum bc \\plus{} 3\\sum(1 \\plus{} a^2 \\minus{} 2bc)^2$\r\n<=>$ 25 \\plus{} 4\\sum b^2c^2 \\plus{} 44abc\\sum \\geq 3\\sum_{cyc} a^4 \\plus{} 40 \\sum bc$\r\nBecause, $ \\sum a^4 \\equal{} 1 \\minus{} 2\\sum b^2c^2$ so the ineq equivalent to:\r\n$ 11 \\plus{} 5\\sum_{cyc} b^2c^2 \\plus{} 22abc\\sum a \\geq 20\\sum bc$\r\nSetting: $ x \\equal{} ab \\plus{} bc \\plus{} ca$\r\nSo we have to prove:$ 11 \\minus{} 20x \\plus{} 5x^2 \\plus{} 12abc\\sum a \\geq 0$\r\nUse schur-inequality :\r\n$ \\sum a^4 \\plus{} 2abc\\sum a \\geq (a^2 \\plus{} b^2 \\plus{} c^2)(ab \\plus{} bc \\plus{} ca)$\r\n=>$ 6abc\\sum a \\geq 2x^2 \\plus{} x \\minus{} 1$\r\n=>$ 11 \\minus{} 20x \\plus{} 5x^2 \\plus{} 12abc\\sum a \\geq 9(x \\minus{} 1)^2 \\geq 0$\r\nDone\r\n\r\n[b]P/s[/b] I haven't checked it yet", + "Solution_4": "Sorry for reviving this old thread but I think I got the nice solution for this inequality!\r\nHere's my solution:\r\nNote that 1/[3+a^2-bc]=1/[2+2a^2+(b-c)^2]\r\nSimilar to 2 more rational polynomial, we will have:LHS=1/[2+2a^2+(b-c)^2]+1/[2+2b^2+(c-a)^2]+1/[2+2c^2+(a-b)^2]\r\nWe see that when we decrease a,b,c by 0==2(a-c)^2,2b^2>=2(b-c)^2 and 2c^2>=0), but the RHS is decreasing(to see it easily, we should rewrite 9/8 by 9(a^2+ b^2+c^2)/8, we will see that: 9(a^2+b^2+c^2)>=9[(a-c)^2+(b-c)^2]\r\nNow, it's sufficient to prove the inequality when one variable is equal to 0. Assume that c=0\r\nThen we need to prove: 1/3+a^2 +1/3+b^2 +1/3-2ab=<9/8(a^2+b^2=1 and a,b>=0).\r\nWe have two ways to solve it from this point!\r\n1st way: Using substitution a^2=1-b^2 and a=square root(1-b^2) and derivative to find the maximum of function f(b)\r\n2nd way: clearing all denominator to prove it(well, a bunch of calculation, though :( ).\r\nThe equality occurs when a=b=c=square root(1/3)(Quite surprising because the entire mixing variable can work here). Anybody can check it for me? :)", + "Solution_5": "In fact, I don't think your solution is right .\r\nFisrt\r\nYou have decrease $ a,b,c$ so $ a^2\\plus{}b^2\\plus{}c^2 \\le 1$\r\nAnd you can't assume $ c\\equal{}0$ .(You can when you have decrease them)\r\n\r\nP/s: I'm not good at English $ \\rightarrow$ Maybe I have some mistakes in grammar or vocabulary .", + "Solution_6": "You're right that a^2+b^2+c^2<=1. But when I decrease all three variables by c=min(a,b,c), then the RHS is still the same since it's a constant! So, the reason you showed is definitely wrong! The only thing that makes me suspicious is about the equality of this inequality. I will try proving:f(a,b,c)<=f[a,square root(b^2+c^2/2).square root(b^2+c^2/2)] since if it's true, then it will show the equality that we need. Otherwise, I will have you find out my mistake in the solution above. :wink:", + "Solution_7": "It is easy with Lagrange Multipliers:\r\nLet $ f(a,b,c)\\equal{}\\dfrac{1}{3\\plus{}a^2\\minus{}2bc}\\plus{}\\dfrac{1}{3\\plus{}b^2\\minus{}2ca}\\plus{}\\dfrac{1}{3\\plus{}c^2\\minus{}2ab}\\leq\\dfrac{9}{8}$ and we will solve the problem only with the condition $ g(a,b,c)\\equal{}a^2\\plus{}b^2\\plus{}c^2\\minus{}1\\equal{}0.$ For the maximum we don't need $ a,b,c\\geq 0$ because it is easily seen that $ f(a,b,c)\\geq f(\\minus{}a,b,c),$ $ f(a,b,c)\\geq f(\\minus{}a,\\minus{}b,c)$ and $ f(a,b,c)\\equal{} f(\\minus{}a,\\minus{}b,\\minus{}c)$ for $ a,b,c\\geq 0.$ Thus the maximum is obtained for $ a,b,c\\geq 0.$\r\nAlso, $ F(a,b,c)\\equal{}f(a,b,c)\\plus{}\\lambda g(a,b,c),$ and solving \\[ \\frac{\\partial F}{\\partial a}\\equal{}0, \\frac{\\partial F}{\\partial b}\\equal{}0, \\frac{\\partial F}{\\partial c}\\equal{}0, \\frac{\\partial F}{\\partial \\lambda}\\equal{}0\\] for $ a,b,c\\geq 0$ we obtain the following solutions: $ a\\equal{}b\\equal{}0,c\\equal{}1, a\\equal{}0,b\\equal{}c$ and $ a\\equal{}b\\equal{}c.$ ($ \\frac{\\partial F}{\\partial a}\\equal{}\\frac{2a(\\lambda \\minus{}1)}{(3\\plus{}a^2\\minus{}2bc)^2}$ so the computations are easy). Hence it is easily seen that the maximum is when $ a\\equal{}b\\equal{}c\\equal{}\\sqrt{3}/3.$", + "Solution_8": "[quote=\"mitdac123\"]:D .It's not enough to prove the first ineq ^^ .But if we use Schur .It's so easy :)\n[b]Solution of chien_than[/b]\n\n<=>$ \\sum_{cyc} \\frac {1 \\plus{} a^2 \\minus{} 2bc}{3 \\plus{} a^2 \\minus{} 2bc} \\geq \\frac {3}{4}$\nUse cauchy-schwarz , we have :\n$ \\sum_{cyc} \\frac {1 \\plus{} a^2 \\minus{} 2bc}{3 \\plus{} a^2 \\minus{} 2bc} \\geq \\frac {4(2 \\minus{} \\sum bc)^2}{8 \\minus{} 4\\sum bc \\plus{} \\sum(1 \\plus{} a^2 \\minus{} 2bc)^2}$\nWe need to prove\n$ 16(2 \\minus{} \\sum bc)^2\\geq 24 \\minus{} 12\\sum bc \\plus{} 3\\sum(1 \\plus{} a^2 \\minus{} 2bc)^2$\n<=>$ 25 \\plus{} 4\\sum b^2c^2 \\plus{} 44abc\\sum \\geq 3\\sum_{cyc} a^4 \\plus{} 40 \\sum bc$\nBecause, $ \\sum a^4 \\equal{} 1 \\minus{} 2\\sum b^2c^2$ so the ineq equivalent to:\n$ 11 \\plus{} 5\\sum_{cyc} b^2c^2 \\plus{} 22abc\\sum a \\geq 20\\sum bc$\nSetting: $ x \\equal{} ab \\plus{} bc \\plus{} ca$\nSo we have to prove:$ 11 \\minus{} 20x \\plus{} 5x^2 \\plus{} 12abc\\sum a \\geq 0$\nUse schur-inequality :\n$ \\sum a^4 \\plus{} 2abc\\sum a \\geq (a^2 \\plus{} b^2 \\plus{} c^2)(ab \\plus{} bc \\plus{} ca)$\n=>$ 6abc\\sum a \\geq 2x^2 \\plus{} x \\minus{} 1$\n=>$ 11 \\minus{} 20x \\plus{} 5x^2 \\plus{} 12abc\\sum a \\geq 9(x \\minus{} 1)^2 \\geq 0$\n[/quote]\r\n\r\n\r\nThis solution is nice, but I found an other Cauchy-Schwartz approach, which is much more elementary, although the type of solution is not new.", + "Solution_9": "[img]http://s5.sinaimg.cn/middle/0018zOAxgy70L8pYG0s34&690[/img]", + "Solution_10": "The following inequality is also true\n[img]http://s16.sinaimg.cn/middle/0018zOAxgy70KJ4kNKT9f&690[/img]" +} +{ + "Tag": [ + "puzzles" + ], + "Problem": "You're an investigator for a murder case and there were 4 people at the scene of the crime, persons A, B, C, D\r\n\r\nYou interogate each person and here's what they say. Each person always lies or always tells the truth.\r\n\r\nA) B always lies and the murderer is not C.\r\nB) C always tells the truth and the murderer is D.\r\nC) D always tells the truth and the murderer is B.\r\nD) A always lies and the murderer is not A.\r\n\r\nWho's the murderer?", + "Solution_1": "Is it B???", + "Solution_2": "[quote=\"shaggy75\"]Is it B???[/quote]\r\n[hide]I got A, starting by assuming A is telling the truth[/hide]", + "Solution_3": "[hide=\"a full solution\"]First, assume A always lies. Then, the murderer would have to be C, and B would have to tell the truth, meaning that the murderer would have to be D, which is a contradiction, so A always tells the truth. This means that B is a liar, and C isn't the murderer. Since B is a liar, C is a liar and D isn't the murderer. Since C is a liar, D is a liar and B isn't the murderer, and this leaves A to be the murderer. Checking to make sure that this works, we know D to be a liar, meaning that A tells the truth, which fits in, and that A is the murderer, which also fits in. Therefore, A IS THE MURDERER!!!!! Ironically, the only person that tells the truth is the murderer, proving that this is one screwed up town.\n[/hide]\r\nYAY! A not spammy post!" +} +{ + "Tag": [ + "geometry", + "ratio", + "AMC" + ], + "Problem": "If circular arcs $AC$ and $BC$ have centers at $B$ and $A,$ then there exists a circle tangent to both arc $AC,BC$ and segment $AB$ If the length of arc $BC$ is 12, then what is the circumference of the circle?", + "Solution_1": "Put this figure into a bigger circle so that the radius of the bigger circle is $AB$, thus using the power point theorem $AF^2=AD(AE)$. Then make $x$ the radius of the small circle then you already know that $AD=AB$ so $AE=AD-ED$ and that's equal to $AB-2x$. So then put that in the first part and you get $\\frac{AB}{2}^2=AB(AB-2x)$ so $x=\\frac{3}{8}(AB)$. Now since $C,B,A$ are the same distance from each other you know that the arc $BC=60$. So you know that the circumference of the circle is $72$ since $BC=12$. Now, here's the thing I might've guessed on, I think you do $\\frac{3}{8}(72)$ because the circumferences are the same ratio as their radii and you get the circumference of the small circle as $27$.", + "Solution_2": "Is my answer right? I believe I've seen this same problem on an old AMC-12, it was like the last question.", + "Solution_3": "yeah, nice solution, 27 was the answer. :surf: i couldnt get the 60 degrees part, so i was confused.", + "Solution_4": "[quote=\"pkothari13\"]yeah, nice solution, 27 was the answer. :surf: i couldnt get the 60 degrees part, so i was confused.[/quote]\r\n\r\nIt was $60$ degrees because if you draw additional lines to it that are tangent to the small circle, you will see that they are all equal, thus the triangle must be equilateral." +} +{ + "Tag": [], + "Problem": "$x+1-\\frac{4}{x+1}$", + "Solution_1": "[hide]Is the answer $(x^2 + 2x - 3) / (x + 1) ?$\n\nYou can change $x + 1$ to $(x^2 + 2x + 1)/(x + 1)$.\n\nFrom there, you just subtract.[/hide]", + "Solution_2": "[quote=\"elementsofmathkid\"][hide]Is the answer $(x^2 + 2x - 3) / (x + 1) ?$\n\nYou can change $x + 1$ to $(x^2 + 2x + 1)/(x + 1)$.\n\nFrom there, you just subtract.[/hide][/quote]\r\nyes that is correct", + "Solution_3": "[quote=\"math92\"]$x+1-\\frac{4}{x+1}$[/quote]\r\n\r\n[hide]$=\\frac{x^2+2x+1}{x+1}-\\frac{4}{x+1}$\n\n$=\\frac{x^2+2x-3}{x+1}$[/hide]", + "Solution_4": "jli, or anybody else. I don't know that much about LaTex. How do you get the fraction bar to be straight, so that you have numbers above and below it?\r\n\r\n\r\nThanks.", + "Solution_5": "elemtsofmathkid, if you put \\frac{2x}{y} surronded by dollar signs you would get $\\frac{2x}{y}$ so basically just put this thing [b]\\[/b], write [b]frac[/b] and put the numerator in these [b]{}[/b] followed by the denominator [b]{}[/b].", + "Solution_6": "Ok. Thanks. Hopefully I will remember that. Sleepy..... :)", + "Solution_7": "[hide]$\\frac{x squared + 2x + 1}{x + 1}$ - $\\frac{4}{x + 1}$ = $\\frac{x squared + 2x - 3}{x +1}$ [/hide]", + "Solution_8": "To do powers, you put x^{power} and subscript is x_{subscript}", + "Solution_9": "[quote=\"nat mc\"]To do powers, you put x^{power} and subscript is x_{subscript}[/quote]\r\nthx for teahing me subscript" +} +{ + "Tag": [ + "geometry", + "3D geometry", + "octahedron", + "probability" + ], + "Problem": "A bug starts at one vertex of a regular octahedron and moves along the edges of the octahedron according to the following rule. At each vertex the bug will choose to travel along one of the four edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after five moves the bug will have visited every vertex exactly once?", + "Solution_1": "This is directly addressed midway through [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125811&start=23]this thread[/url]. The URL is such that you will start from the first relevant post of that thread.\r\n\r\n[hide=\"Alternately\"]Pick a face and label the three vertices A, B, C. Label the other three vertices the same as the one directly opposite them, so that each A borders 2 B's and 2 C's, etc.\n\nNow, let the starting vertex be one of the A's. Find the probability that you will be at a vertex A in $ n$ moves, and then divide by 2 because after the first move, both A's are symmetric.\n\nSo now, you can biject to the problem of a bug moving along an equilateral triangle, which is exactly the original problem of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125811]the same thread I linked above[/url].[/hide]" +} +{ + "Tag": [ + "algebra", + "polynomial", + "search", + "combinatorics proposed", + "combinatorics" + ], + "Problem": "Let $ n \\ge 1$ and $ k \\ge 3$ be integers. A circle is divided into $ n$ sectors $ a_1,a_2,\\dots,a_n$. We will color the $ n$ sectors with $ k$ different colors such that $ a_i$ and $ a_{i \\plus{} 1}$ have different color for each $ i \\equal{} 1,2,\\dots,n$ where $ a_{n \\plus{} 1}\\equal{}a_1$. Find the number of ways to do such coloring.", + "Solution_1": "You are looking for the chromatic polynomial of the cycle.\r\nHave a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=1545682589&t=16046\r\n\r\nThe correct formula is at post #6 (even if the reasoning is at #2).\r\n\r\nPierre." +} +{ + "Tag": [ + "inequalities", + "inequalities unsolved" + ], + "Problem": "If $a,b,c>0$ prove that:\r\n$\\frac{a^{3}}{a^{2}+ab+b^{2}}+\\frac{b^{3}}{b^{2}+bc+c^{2}}+\\frac{c^{3}}{c^{2}+ca+a^{2}}\\geq \\frac{a+b+c}{3}$", + "Solution_1": "Posted before. In that, we prove $\\frac{a^{3}}{a^{2}+ab+b^{2}}\\geq\\frac{2a-b}{3}$, e.t.c.", + "Solution_2": "or , take a look at the sum \r\n$B=\\sum \\frac{b^{3}}{a^{2}+b^{2}+ab}$\r\nwe can easily see that $A-B=a-b+b-c+c-a = 0$ so $A=B$\r\nand then $A+B=\\sum\\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}$\r\nwe will prove that $\\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}\\geq \\frac{a+b}{3}$ , its equivalent to $a^{3}+b^{3}\\geq a^{2}b+ab^{2}$ which is true , so our proof is complete", + "Solution_3": "I am sorry but I did not know that it was pposted before :blush: \r\nTank you for your nice solutions :wink:" +} +{ + "Tag": [ + "probability", + "function", + "integration", + "calculus", + "parameterization", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Hello, I have two questions in probabilities :\r\n\r\n1/ A boy is looking outside the window and is counting $X(d)$ - the number of yellow cars he can see during a time $d$. Suppose the probability that the number of yellow cars he sees during $d$ seconds is $k$ is $e^{-\\alpha d}\\frac{(\\alpha d)^{k}}{k!}$.\r\n\\[P[X(d)=k]=e^{-\\alpha d}\\frac{(\\alpha d)^{k}}{k!}\\]\r\nAlso, it is known that if $T$ is the time he has to wait to see a red car is such that\r\n\\[P[T>t]=e^{-\\beta t}. \\]\r\nShow that if $Y$ is the number of yellow cars the boy will see before e sees a red car is such that\r\n\\[P[Y=k]=[\\frac{\\alpha}{\\alpha+\\beta}]^{k}\\frac{\\beta}{\\alpha+\\beta}\\]\r\n2/ Suppose you have 2000 balls and 100 holes. Someone put randomly all the balls in the holes. I'm before a hole. What is the probability that in this hole there are exactly k balls ?\r\n\r\nThanks", + "Solution_1": "2/ there are $100^{2000}$ ways to put all balls into holes. This is the number of all possible outcomes. How many outcomes are favorable to you? Choose $k$ balls out of $2000$ (can be done in $\\binom{2000}{k}$ ways), put them into the hole you are looking at; and then put the rest into the remaining holes (can be done in $99^{2000-k}$ ways). Thus, $\\binom{2000}{k}99^{2000-k}$ out of $100^{2000}$ outcomes are favorable, which gives the probability of $\\binom{2000}{k}\\frac{99^{2000-k}}{100^{2000}}$", + "Solution_2": "The probability distributions of problem 1 match the so-called \"Poisson process.\" There are some very strong assumptions of independence being made here - sightings in any interval are independent of those in any nonoverlapping interval, and we must assume that sightings of red cars in any interval are independent of sightings of yellow cars in that same interval.\r\n\r\nHowever, the statement about the continuous random variable $T$ is incorrect. That is not $P(T=t),$ since $P(T=t)=0$ for all $t.$ That function, $\\beta e^{-\\beta t}$ for $t>0,$ is actually the probability density function for $T.$ \r\n\r\nIt's given that the number of yellow cars in a time interval $d$ is a Poisson distribution of mean $\\alpha d$ an that the waiting time until a red car is an exponential distribution of mean $1/\\beta.$ Both of those descriptions can be flipped around: the number of red cars in a time interval $d$ is a Poisson distribution of mean $\\beta d$ and the waiting time until a yellow car is an exponential distribution of mean $1/\\alpha.$\r\n\r\nLet $U$ equal the time until the sighting of the first yellow car, and let $W$ be the time until the sigting of the first red or yellow car. $P(T>t)=e^{-\\beta t}$ and $P(U>t)=e^{-\\alpha t}.$ $W=\\min(U,T),$ and, with the assumption that $U$ and $T$ are independent, we get that $P(W>t)=P(U>t)\\cdot P(T>t)=e^{-(\\alpha+\\beta)t}$ for $t>0.$ That makes $W$ an exponentially distributed random variable. \r\n\r\nIgnore all cars that are neither red nor yellow (which is what I have already been doing.) What is the probability that first car seen is red - that is $P(T=W)=P(T 0$ there exists a square of side-length $L$ with side parallel to the axes of ccordinates such that any point in the interior of the square or on its boundary is invisible.\r\nThis will lead to the desired result since, for any given $R$, we may choose $L$ sufficientely large so that the square covers a disk with radius $R$.\r\n\r\nClearly, the lattice point $M(a,b)$ is invisible if and only if $\\gcd (a,b) > 1$.\r\n\r\nNow, let $L > 0$ be given.\r\nLet $p_{i,j}$ be pairwise distinct primes for $i=0, \\cdots, L$ and $j = 0 , \\cdots , L$.\r\nFrom the chinese remainders theorem, we know that there exists $x,y > 0$ such that $x =- i$ mod[$p_{i,j}$] and $y=-j$ mod[$p_{i,j}$], for all $i,j$.\r\n\r\nThen we choose the point $M(x,y)$ to be the lower left corner of our desired square, and wed are done." +} +{ + "Tag": [ + "induction", + "strong induction" + ], + "Problem": "Hey guys, I just started to work on some algebra and proves and I'm working on this problem that I can't figure out. Prove that for n>=2, if n^2-4 is a prime number, then n = 3. i'm not sure how to start it off...any ideas? Thankz for your help!", + "Solution_1": "[hide=\"Hint\"]\nThe key thing to notice is that $4$ is a square, so the equation is of the form $a^{2}-b^{2}$.[/hide]\n\n[hide=\"Solution\"]\n$n^{2}-4=(n+2)(n-2)$, so for all $n$, $(n-2)|(n^{2}-4)$.\nFor $n>3$, $(n-2)>1$, so this implies that $n^{2}-4$ is composite.[/hide]", + "Solution_2": "ty so much!", + "Solution_3": "[hide=\"My Solution\"]\n\nIf $n^{2}-4$ is prime, then $(n+2)(n-2)$ is also prime. For this to be prime, either one of them must be one, and the other the prime. If $n+2=1$, $n=-1$ which wouldn't make sense. Then, we know that $n-2=1$ which yields $n=3$.[/hide]\n\n[hide=\"Huge hint\"]\nFactor[/hide]", + "Solution_4": "ahh.... you guys are so smart :lol: \r\n\r\nI've stumbled onto another problem. It's kind of long though :wink: \r\n\r\nA pile starts with n marbles, with n>=2. THe pile is split into two piles, and the product of the sizes of the piles is calculated. One of these new piles is then split into two piles, with the product of the sizes of these two new piles calculated added to the ( sum of) the previous product(s). This process continues until we have n piles with 1 marble each. For example, starting with 5 marbles,we could split 2,3 (giving 6), then 2,2,1(giving 6+2) then 1,1,2,1 (giving 6 + 2+1), then 1,1,1,1,1 giving 6+2 +1 +1=10. Prove using strong induction that, no matter how the pile is split, the total at the end is always (1/2)(n^2-n)", + "Solution_5": "You should put this in a different thread, instead of turning this into a marathon. Also use latex.b", + "Solution_6": "Let us call the total of all the products after the process has been finished $P(n)$. Note that $P(2)=\\frac{2^{2}-2}{2}$. Suppose that $P(m)=\\frac{m^{2}-m}{2}$ for all $m 0$ then prove that:\r\n$ \\sum \\frac {a^3}{1 \\minus{} a^8} \\ge \\frac {9 \\sqrt [4]{3}}{8}$\r\n(sorry,i made a stupid mistake,should be moved to: inequality,proposed and own problems forum :oops: )", + "Solution_1": "$ \\frac {a^3}{1 \\minus{} a^8} \\ge \\frac {9.\\sqrt [4]{3}.a^4}{8}$\r\n$ \\Longleftrightarrow a^9 \\plus{} \\frac {8}{9\\sqrt [4]{3}} \\ge a$\r\n\r\nwhich is Am-Gm since \r\n\r\n$ a^9 \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\plus{} \\frac {1}{9\\sqrt [4]{3}} \\ge a$", + "Solution_2": "i saw this pro blem a little different:find the minimum value of a^3/1-a^8 and cyc but i doesn't matter:\r\nWe have that a^4/a(1-a^8) attains it's minimal value when it's denominator is in its maximum, so let's see when a(1-a^8) is maximum. Let k be this maximum. We have that\r\n 8a^8a(1-a^8)^8=<8k^8.By a.m-g.m \r\nwe get that [a^8+(1-a^8)+1+1+1+1+1+1+1/9]^9>= 8a^8/a(1-a^8)>= 8a^8a(1-a^8)^8\r\n(8/9)^9>= 8a^8a(1-a^8)^8\r\n8k^8=(8/9)^9 and k=8/9sqrt(4)3 then a^4/a(1-a^8)>=a^4 9sqrt(4)3/8 and cyc then sum up the result since the answer:9sqrt(4)3" +} +{ + "Tag": [ + "inequalities", + "inequalities unsolved" + ], + "Problem": "For $ a,b,c\\geq \\minus{}\\frac{3}{4}$ and $ a\\plus{}b\\plus{}c\\equal{}1$. Prove:\r\n$ \\frac{a}{a^{2}\\plus{}1}\\plus{}\\frac{b}{b^{2}\\plus{}1}\\plus{}\\frac{c}{c^{2}\\plus{}1}\\leq \\frac{9}{10}$", + "Solution_1": "$ (3a\\minus{}1)^2(4a\\plus{}3) \\geq 0$\r\n\r\nthus\r\n\r\n$ 36a^3\\plus{}3a^2\\minus{}14a\\plus{}3 \\geq 0$\r\n\r\nand therefore\r\n\r\n$ 3(a^2\\plus{}1)(12a\\plus{}1) \\geq 50a$\r\n\r\nnow we have $ \\frac{a}{a^2\\plus{}1} \\leq \\frac{36a\\plus{}3}{50}$, add the other two inequalities up and then the result follows. :)" +} +{ + "Tag": [ + "MATHCOUNTS" + ], + "Problem": "Everybody get ready for the Math Challenger.\r\nFor Math Challenger, the rules are answer the question first!\r\n QUESTION:\r\n\r\n Given that 6x + y = 15, the value of 3x can be written in terms of y as ay+b for some numbers a and b. What is the simplified value of a+b?", + "Solution_1": "we have 6 for the coeficient of x. we want that to be 3. so divide by 2. we have $ 3x\\plus{}\\frac{1}{2}y\\equal{}\\frac{15}{2}$ so $ x\\equal{}\\frac{15}{2}\\minus{}\\frac{1}{2}y$ we have 15/2-1/2=14/2=7\r\n\r\n\r\n[hide][size=200][color=red][i][b][u]7[/u][/b][/i][/color][/size][/hide]", + "Solution_2": "Ehhh, I knew abacadaea got that first:\r\nif 6x+y=15, then 6x=15-y, 3x=7.5-0.5y, so a+b=7.5-0.5=7.", + "Solution_3": "I think we already have the Mathcounts Problem Marathon... 2!, and there's no need for another marathon. Locked." +} +{ + "Tag": [ + "AMC", + "AIME" + ], + "Problem": "Would someone be so kind as to explain the solution for this problem:\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_8\r\n\r\nto me?\r\n\r\nWhy must our two digit prime p satisfy 3p<200?", + "Solution_1": "I think its because their needs to be 3 multiples in the numerator in order to cancel out the denominator and leave one multiple of that prime in the numerator. Is this right?\r\n\r\nThere are 3 multiples of 61 in the numerator and only 2 in the denominator. So the two 61s cancel with the ones in the denominator leaving 61 as the largest prime on top.", + "Solution_2": "I'm not entirely sure on this, but, since the prime factor is in $ 100!$ once (if its greater than 50), there will be two of that factor in the denominator, so there must be three or more in the numerator, which gives $ 3p<200$.", + "Solution_3": "Thank you. Both of your explanations are very clear. \r\nFor some reason, this AIME problem solution seems too... short to be a number 8.", + "Solution_4": "I really think this stuff on AIME should be in the AMC forum. I mean, it's about stuff from one of the AIME contests on American Mathematics Competitions." +} +{ + "Tag": [ + "AMC", + "USA(J)MO", + "USAMO", + "function" + ], + "Problem": "4. Let P be a convex polygon with n sides, n \u00b8 3. Any set of n\u00a13 diagonals of P that do not\r\nintersect in the interior of the polygon determine a triangulation of P into n\u00a12 triangles.\r\nIf P is regular and there is a triangulation of P consisting of only isosceles triangles, \u00afnd\r\nall the possible values of n.", + "Solution_1": "[url=http://www.artofproblemsolving.com/Wiki/index.php/2008_USAMO_Problems/Problem_4]AoPS Wiki[/url]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202905]Previous Topic[/url]", + "Solution_2": "I actually want to see someone's ORIGINAL solution.", + "Solution_3": "those are ORIGINAL solutions...they were the ones that they came up with while taking the test....", + "Solution_4": "My solution in the previous thread is only slightly cleaned up. That's essentially what I wrote on the test, except that I defined a second function that I later realized I didn't need. The official solution is even nicer." +} +{ + "Tag": [ + "function", + "calculus", + "derivative", + "parameterization", + "trigonometry", + "inequalities", + "limit" + ], + "Problem": "Let z complex number such that 2|z|=|z-1|. Find the maximum value of |z+(1/z)|.", + "Solution_1": "Let $ z = a + bi$ where $ a,b \\in R$.\r\nWe use the equation $ 2|z| = |z - 1|$ to express $ a^2 + b^2$ in terms of $ a$.\r\nSquaring we get\r\n$ 4|z|^2 = |z - 1|^2$\r\n$ 4|a + bi|^2 = |a - 1 + bi|^2$\r\n$ 4(a^2 + b^2) = (a - 1)^2 + b^2$\r\n$ 4a^2 + 4b^2 = a^2 - 2a + 1 + b^2$\r\n$ 3a^2 + 3b^2 = 1 - 2a$\r\n$ a^2 + b^2 = \\dfrac{1 - 2a}{3}$\r\nAlso, we find the range of values for $ a$.\r\n$ 9a^2 + 9b^2 = 3 - 6a$\r\n$ 9a^2 + 6a + 1 + 9b^2 = 4$\r\n$ (3a + 1)^2 + (3b)^2 = 4$\r\n$ (3a + 1)^2 \\le 4$\r\n$ (3a - 1)(3a + 3) \\le 0$\r\n$ - 1 \\le a \\le \\dfrac{1}{3}$\r\nWe express $ |z + \\dfrac{1}{z}|^2$ as a function of $ a$.\r\n$ |z + \\dfrac{1}{z}|^2 = |a + bi + \\dfrac{1}{a + bi}|^2$\r\n$ = |a + bi + \\dfrac{a-bi}{a^2 + b^2}|^2$\r\n$ = |a(1 + \\dfrac{1}{a^2 + b^2}) + bi(1 - \\dfrac{1}{a^2 + b^2})|$\r\n$ = a^2(1 + \\dfrac{1}{a^2 + b^2})^2 + b^2(1 - \\dfrac{1}{a^2 + b^2})^2$\r\n$ = a^2 + b^2 + \\dfrac{1 + 4a^2}{a^2 + b^2} - 2$.\r\nSubstituting in $ a^2 + b^2 = \\dfrac{1 - 2a}{3}$ we get\r\n$ |z + \\dfrac{1}{z}|^2 = \\dfrac{1 - 2a}{3} + \\dfrac{1 + 4a^2}{(\\frac {1 - 2a}{3})} - 2 = - \\dfrac{14}{3} - \\dfrac{20 a}{3} - \\dfrac{6}{2 a - 1}$.\r\nWe show the convexity of $ |z + \\dfrac{1}{z}|^2$.\r\nTaking the second derivative, the first and second terms vanish, and we have\r\n$ \\dfrac{d^2}{da^2}(|z + \\dfrac{1}{z}|^2) = \\dfrac{d^2}{da^2}\\left( - \\dfrac{6}{2 a - 1}\\right) = \\dfrac{d}{da}\\left(\\dfrac{12}{(2a - 1)^2}\\right) = \\dfrac{ - 48}{(2a - 1)^3}$.\r\n$ 2a - 1 < 0$ because $ - 1 \\le a \\le \\dfrac{1}{3} < \\dfrac{1}{2}$, so $ (2a - 1)^3 < 0$, implying $ \\dfrac{d^2}{da^2}|z + \\dfrac{1}{z}|^2 = \\dfrac{ - 48}{(2a - 1)^3} > 0$.\r\nTherefore, $ |z + \\dfrac{1}{z}|^2$ is convex. $ |z + \\dfrac{1}{z}|^2$ cannot have a local maximum. It can only have an absolute maximum. Thus we check the boundaries of $ a$.\r\nWhen $ a = - 1$\r\n$ |z + \\dfrac{1}{z}|^2 = - \\dfrac{14}{3} - \\dfrac{20 ( - 1)}{3} - \\dfrac{6}{2 ( - 1) - 1} = 4$.\r\nWhen $ a = \\dfrac{1}{3}$\r\n$ |z + \\dfrac{1}{z}|^2 = - \\dfrac{14}{3} - \\dfrac{20 (\\frac {1}{3})}{3} - \\dfrac{6}{2 (\\frac {1}{3}) - 1} = \\dfrac{100}{9}$.\r\nThe maximum of $ |z + \\dfrac{1}{z}|^2$ is $ \\dfrac{100}{9}$.\r\nTherefore, the maximum value of $ |z + \\dfrac{1}{z}|$ is $ \\sqrt \\dfrac{100}{9} = \\dfrac{10}{3}$. It occurs at $ z = \\dfrac{1}{3}$: \r\n$ |z + \\dfrac{1}{z}| = |\\dfrac{1}{3} + 3| = \\dfrac{10}{3}$.", + "Solution_2": "Mij, you have misidentified the set $ \\{z: 2|z|\\equal{}|z\\minus{}1|\\}.$ That set is a circle.\r\n\r\nWrite the equation as\r\n\r\n$ 4|z|^2\\equal{}|z\\minus{}1|^2$\r\n\r\n$ 4z\\overline{z}\\equal{}(z\\minus{}1)\\overline{(z\\minus{}1)}\\equal{}z\\overline{z}\\minus{}\\overline{z}\\minus{}z\\plus{}1$\r\n\r\n$ 3z\\overline{z}\\plus{}\\overline{z}\\plus{}z\\equal{}1$\r\n\r\n$ z\\overline{z}\\plus{}\\frac13\\,\\overline{z}\\plus{}\\frac13\\,z\\plus{}\\frac19\\equal{}\\frac13\\plus{}\\frac19$\r\n\r\n$ \\left(z\\plus{}\\frac13\\right)\\overline{\\left(z\\plus{}\\frac13\\right)}\\equal{}\\frac49$\r\n\r\n$ \\left|z\\plus{}\\frac13\\right|^2\\equal{}\\frac49$\r\n\r\n$ \\left|z\\plus{}\\frac13\\right|\\equal{}\\frac13.$\r\n\r\nThis is a circle centered at $ \\minus{}\\frac13$ with radius $ \\frac23.$ It intersects the real axis at two points, $ \\minus{}1$ and $ \\frac13,$ but remember that it is a circle.\r\n\r\nThe function $ f(z)\\equal{}z\\plus{}\\frac1z\\equal{}\\frac{z^2\\plus{}1}{z}$ is continuous on that circle, and the circle is a compact set. $ f$ will assume both an absolute maximum and an absolute minimum on that set.\r\n\r\nOne possible approach: write $ z$ on the circle as $ z\\equal{}\\frac{\\minus{}1\\plus{}2e^{i\\theta}}3$ for $ 0\\le\\theta\\le 2\\pi.$ That's a parameterization of the circle. Then you can write the thing to be maximized in terms of $ \\theta.$ It will be a rational expression in sines and cosines of $ \\theta.$ (It may still be messy, but that's a start.)", + "Solution_3": "I do not see a flaw in my proof. I have shown that the value of $ a^2 \\plus{} b^2$ depends only on $ a$ (whatever $ b$ is) and that the value of $ |z \\plus{} \\dfrac{1}{z}|^2$ depends only on $ a$ (whatever $ b$ is) because of the condition $ 2|z| \\equal{} |z \\minus{} 1|$.\r\nIf I let $ z \\equal{} \\frac { \\minus{} 1 \\plus{} 2e^{i\\theta}}{3}$ and graph the function\r\n$ |z \\plus{} \\dfrac{1}{z}| \\equal{} |\\frac { \\minus{} 1 \\plus{} 2e^{i\\theta}}{3} \\plus{} \\frac {3}{ \\minus{} 1 \\plus{} 2e^{i\\theta}}|$ over $ 0 \\le \\theta \\le 2\\pi$ (I used Mathematica),\r\nI get the maximum $ \\dfrac{10}{3}$ at $ 0$ and $ 2\\pi$. That corresponds to\r\n$ z \\equal{} \\frac { \\minus{} 1 \\plus{} 2e^0}{3} \\equal{} \\dfrac{1}{3}$.", + "Solution_4": "Mij's result is right. here is a little bit shorter demo :\r\n\r\nUsing $ z\\equal{}\\rho e^{i\\theta}$, the condition $ 2|z|\\equal{}|z\\minus{}1|$ becomes $ 4\\rho^2\\equal{}(\\rho\\cos(\\theta)\\minus{}1)^2\\plus{}(\\rho\\sin(\\theta))^2$, so $ 3\\rho^2\\plus{}2\\rho\\cos(\\theta)\\minus{}1\\equal{}0$\r\n\r\nThis condition gives $ \\cos(\\theta)\\equal{}\\frac{1\\minus{}3\\rho^2}{2\\rho}$ with the condition $ |\\frac{1\\minus{}3\\rho^2}{2\\rho}|\\leq 1$\r\n\r\nThe last inequality may be written $ (1\\minus{}3\\rho^2)^2\\leq 4\\rho^2$ or again $ 9\\rho^4\\minus{}10\\rho^2\\plus{}1\\leq 0$ or again $ (\\rho^2\\minus{}1)(9\\rho^2\\minus{}1)\\leq 0$, so $ \\rho^2\\in[\\frac{1}{9},1]$\r\n\r\nSo the equality $ 2|z|\\equal{}|z\\minus{}1|$ is equivalent to $ \\rho\\in[\\frac{1}{3},1]$ and $ 2\\cos(\\theta)\\equal{}\\frac{1}{\\rho}\\minus{}3\\rho$\r\n\r\nThen $ A\\equal{}|z\\plus{}\\frac{1}{z}|^2\\equal{}|\\rho e^{i\\theta}$ $ \\plus{}\\frac{1}{\\rho}e^{\\minus{}i\\theta}|^2$ $ \\equal{}(\\rho\\plus{}\\frac{1}{\\rho})^2\\cos^2(\\theta)$ $ \\plus{}(\\rho\\minus{}\\frac{1}{\\rho})^2\\sin^2(\\theta)$ $ \\equal{}\\rho^2 \\plus{} \\frac{1}{\\rho^2} \\plus{} (2\\cos(\\theta))^2 \\minus{}2$\r\n\r\nUsing then $ 2\\cos(\\theta)\\equal{}\\frac{1}{\\rho}\\minus{}3\\rho$, we get $ A\\equal{}\\rho^2 \\plus{} \\frac{1}{\\rho^2} \\plus{} (\\frac{1}{\\rho}\\minus{}3\\rho)^2 \\minus{}2$ $ \\equal{}10\\rho^2\\plus{}\\frac{2}{\\rho^2}\\minus{}8$ with $ \\rho\\in[\\frac{1}{3},1]$\r\n\r\n$ A(\\rho)$ is a convex function of $ \\rho$ on $ \\mathbb R^\\plus{}$ (as sum of elementary convex functions) and reaches its maximum over $ [\\frac{1}{3},1]$ either for $ \\rho\\equal{}\\frac{1}{3}$, either for $ \\rho\\equal{}1$.\r\n\r\n$ A(\\frac{1}{3})\\equal{}\\frac{100}{9}$ and $ A(1)\\equal{}4$\r\n\r\nSo the maximum of $ |z\\plus{}\\frac{1}{z}|$ is $ \\sqrt A\\equal{}\\frac{10}{3}$ and is reached for $ \\rho\\equal{}\\frac{1}{3}$ and $ \\cos(\\theta)\\equal{}\\frac{1\\minus{}3\\rho^2}{2\\rho}$ $ \\equal{}1$, so for $ z\\equal{}\\frac{1}{3}$, which is Mij's result", + "Solution_5": "Another approach:\r\n\r\n[i]1- [/i]$ 2|z| \\equal{} |z \\minus{} 1| \\implies |z \\plus{} \\frac {1}{3}| \\equal{} \\frac {2}{3}$.\r\n[i]2- [/i]By triangle inequality we have $ |z| \\minus{} | \\minus{} \\frac {1}{3}|\\le |z \\minus{} ( \\minus{} \\frac {1}{3})|\\implies \\frac {1}{3}\\le |z| \\le 1$.\r\n[i]3-[/i] If we fix a positive real $ a$, consider locus of point of $ \\mathbb{C}$ such that $ |z \\plus{} \\frac {1}{z}| \\equal{} a$ that is equivalent to $ \\frac {|z|^4 \\plus{} (z \\plus{} \\overline{z})^2 \\minus{} 2|z|^2 \\plus{} 1}{|z|^2} \\equal{} a^2$. But it implies $ |z|^4 \\minus{} |z|^2(a^2 \\plus{} 2) \\plus{} 1 \\equal{} \\minus{} (z \\plus{} \\overline{z})^2 \\le 0 \\implies \\frac { \\minus{} a \\plus{} \\sqrt {a^2 \\plus{} 4}}{2} \\le |z| \\le \\frac {a \\plus{} \\sqrt {a^2 \\plus{} 4}}{2}$. We are interested to maximize $ a$, but lhs is a decreasing function and rhs a incresing function in $ \\mathbb{R}^ \\plus{}$. So it sufficies to take $ |z|$ minimum possible ( it is possible only in $ z \\equal{} \\frac {1}{3}$ as Patrick rightly said).", + "Solution_6": "[quote=\"bboypa\"]Another approach:\n\n[i]1- [/i]$ 2|z| \\equal{} |z \\minus{} 1| \\implies |z \\plus{} \\frac {1}{3}| \\equal{} \\frac {2}{3}$.[/quote]\nOK. This is Kent's remark\n[quote=\"bboypa\"][i]2- [/i]By triangle inequality we have $ |z| \\minus{} | \\minus{} \\frac {1}{3}|\\le |z \\minus{} ( \\minus{} \\frac {1}{3})|\\implies \\frac {1}{3}\\le |z| \\le 1$.[/quote]\nOK\n[quote=\"bboypa\"][i]3-[/i] If we fix a positive real $ a$, consider locus of point of $ \\mathbb{C}$ such that $ |z \\plus{} \\frac {1}{z}| \\equal{} a$ that is equivalent to $ \\frac {|z|^4 \\plus{} (z \\plus{} \\overline{z})^2 \\minus{} 2|z|^2 \\plus{} 1}{|z|^2} \\equal{} a^2$. But it implies $ |z|^4 \\minus{} |z|^2(a^2 \\plus{} 2) \\plus{} 1 \\equal{} \\minus{} (z \\plus{} \\overline{z})^2 \\le 0 \\implies \\frac { \\minus{} a \\plus{} \\sqrt {a^2 \\plus{} 4}}{2} \\le |z| \\le \\frac {a \\plus{} \\sqrt {a^2 \\plus{} 4}}{2}$. [/quote]\nOK.\n[quote=\"bboypa\"]We are interested to maximize $ a$, but lhs is a decreasing function and rhs a incresing function in $ \\mathbb{R}^ \\plus{}$. So it sufficies to take $ |z|$ minimum possible ( it is possible only in $ z \\equal{} \\frac {1}{3}$ as Patrick rightly said).[/quote]\r\n\r\nHere, I dont understand how you conclude from $ \\frac { \\minus{} a \\plus{} \\sqrt {a^2 \\plus{} 4}}{2} \\le |z| \\le \\frac {a \\plus{} \\sqrt {a^2 \\plus{} 4}}{2}$ that you must take $ |z|$ minimum ....", + "Solution_7": "Do'h what a silly mistake :o Thanks Patrick, it is also easier than solve the previous quadratic..\r\n\r\n$ \\max\\{|z \\plus{} \\frac {1}{z}|\\}$ iff $ \\max\\{\\frac {|z|^4 \\plus{} (z \\plus{} \\overline{z})^2 \\minus{} 2|z|^2 \\plus{} 1}{|z|^2}\\}$ as before, but $ z \\plus{} \\overline{z} \\equal{} 1\\minus{}3|z|^2$ (point1) so it equivalent to maximise a expression like $ f(y) \\equal{} 5y \\plus{} \\frac {1}{y}$ where $ y: \\equal{} |z|^2$. But we know the shape of this curve and by am-gm it has minimum in $ y \\equal{} \\sqrt {5}$, so it is decreasing before $ 1$ and in $ z \\equal{} \\frac {1}{3}$ we have the max (it is unique also for geometric reason..). Thanks again!\r\n\r\n\r\nEdit @pco: I have solved it at mind in the noise of the subway, i was not taking care at sign, because i will square it :roll: however i have edited :wink:", + "Solution_8": "[quote=\"bboypa\"]... but $ z \\plus{} \\overline{z} \\equal{} 3|z|^2 \\minus{} 1$ (point1) ... [/quote]\r\n\r\nJust a little correction : $ z \\plus{} \\overline{z} \\equal{} 1 \\minus{} 3|z|^2$ :)", + "Solution_9": "[quote=\"Kent Merryfield\"]The function $ f(z) \\equal{} z \\plus{} \\frac1z \\equal{} \\frac {z^2 \\plus{} 1}{z}$ is continuous on that circle, and the circle is a compact set. $ f$ will assume both an absolute maximum and an absolute minimum on that set.[/quote]\r\nI don't understand how a function can be continous on a circle. (That is: I don't understand what i means)\r\nI have learned that a function $ f(z)$ i continous whenever $ \\lim_{z \\to z_0} f(z) \\equal{} f(z_0)$.\r\nMy guess as to what it means: If $ g(t) : I \\to \\mathbb{C}$ is a (smooth) parametrisation of the circle (e.g. $ g(t) \\equal{} \\frac {2e^{it} \\minus{} 1}{3}$, where $ I \\equal{} [0;2\\pi)$) then $ \\lim_{t \\to t_0} f(g(t)) \\equal{} f(g(t_0))$. Smooth implies continuity so it should be pretty obvious if that's the case :)\r\nFurther more: What is a compact set? And how do these two facts imply that $ f$ will assume both an absoltu max and min on this compact set?\r\n\r\nA lot of questions, but I hope you have the time to answer them, although they probably are trivial to you.\r\n\r\n(Btw. It should be $ \\left|z \\plus{} \\frac {1}3\\right| \\equal{} \\frac {2}3.$ and not $ \\left|z \\plus{} \\frac {1}3\\right| \\equal{} \\frac {1}3.$ in your post)", + "Solution_10": "Hmm.... $ \\left | z \\plus{} \\frac {1}{z} \\right | \\equal{} \\left | \\frac {z^2 \\plus{} 1}{z} \\right | \\le \\frac {|z|^2 \\plus{} 1}{|z|}$. Drawing the circle, we easily conclude that $ \\frac {1}{3} \\le |z| \\le 1$. But the function $ |z| \\to \\frac {|z|^2 \\plus{} 1}{|z|}$ is increasing in $ (0;1)$, $ \\left | z \\plus{} \\frac{1}{z} \\right | \\le \\frac {|z|^2 \\plus{} 1}{|z|} \\le \\frac{10}{3}$ with equality only if $ |z| \\equal{} \\frac {1}{3}$. Looking at the drawing, $ z \\equal{} \\frac {1}{3}$ is obviously the only one lying on the circle with $ |z| \\equal{} \\frac {1}{3}$, and it luckily satisfies $ \\left | z \\plus{} \\frac {1}{z} \\right | \\equal{} \\frac {10}{3}$. So the answer is $ z \\equal{} \\frac {1}{3}$. (The only reason the solution got so short, was that I knew the answer already..)" +} +{ + "Tag": [], + "Problem": "Okay,this is my [b][size=150]1337[/size][/b]th post and since there is something weird n special about it,i'm going to keep it for the longest time in the history of aops....put ur bets before this gets deleted :D", + "Solution_1": "Well, seeing the time and the date, chances are more likely that it WON'T be deleted very quickly. I know that Miths Apprentice is going to delete this topic, because it is useless, so I will only post this post in it. Nothing more.", + "Solution_2": "Since i won't get a prize or something for this,i'm not going to do it anymore....", + "Solution_3": "Yes, mb. is right. No pint in posting in a thread which will be deleted soon." +} +{ + "Tag": [ + "logarithms", + "calculus", + "calculus computations" + ], + "Problem": "prove that 2^x +3^x +6^x =x^2 has only one root (-1)", + "Solution_1": "For $ x < 0$ the left hand side is strictly increasing while the right hand side is strictly decreasing, so there is at most one root with $ x < 0$. For $ x\\geq 0$, by comparison to the Taylor series of $ 6^x$ we have $ 6^x > 1 \\plus{} x\\log 6 \\plus{} x^2 \\frac {(\\log 6)^2}{2} > x^2$, so the left hand side is strictly greater than the left hand side for all $ x\\geq 0$.", + "Solution_2": "thankyou very much" +} +{ + "Tag": [ + "search" + ], + "Problem": "What is with the name of under each person's username? For example, mine is something like Hodge Conjecture.\r\n\r\nCan you change it? What does it mean?", + "Solution_1": "it's one of the unsolved millenium problems, or something like that. it changes according to the number of your posts on the forum.", + "Solution_2": "It is one of the unsolved millenium problems. I have a feeling that this has been discussed in the forum already (do a search)\r\n\r\nHere is a link for more information\r\n[url=http://www.claymath.org/millennium/]Millenium Problems[/url]\r\n\r\nMy suspicions were correct: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Millenium+problems&t=24794]Forum talk[/url]", + "Solution_3": "is it about your total posts or the posts you sent daily", + "Solution_4": "It's about total posts." +} +{ + "Tag": [ + "calculus", + "derivative", + "function", + "Support", + "topology", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Consider omega, an open set of R^n with smooth boundary.\r\n\r\nLet u be a function in H^2 intersection H^1, which goes to 0 on the boundary of omega. Let v_n be a sequence of functions in C^infinity with compact support, approximating u in H^1(omega), and let w_n be a sequence of functions in C^infinity approximating u in H^2(omega closure). \r\n\r\nDoes it possibly follow that D^2(v_n - w_n) forms a Cauchy sequence in L^2?", + "Solution_1": "Consider the case $u\\equiv 0$ and $w_{n}\\equiv 0$. You ask if $v_{n}\\to 0$ in $H^{1}$ implies that $(D^{2}v_{n})$ is Cauchy (=convergent) in $L^{2}$.\r\n\r\nNaturally, the answer is no: we have no control over the second derivatives of $v_{n}$." +} +{ + "Tag": [ + "induction", + "calculus", + "integration", + "algebra", + "polynomial", + "combinatorics proposed", + "combinatorics" + ], + "Problem": "a) Prove that for any integer $ k>1$, there exist $ k$ distinct integers $ a_1,a_2,\\cdots,a_k$ such that $ a_n\\plus{}a_m|a_na_m$ for any $ 1\\leq n,m\\leq k$.\r\nb) Consider the same problem as above with the divisibility conditioned changed to $ a_n\\plus{}a_m\\plus{}1 | a_na_m$. Does the problem statement hold in this case? (I couldn't solve this one!)", + "Solution_1": "the first problem...\r\n\r\n we're going to give an inductive construction. start with the set $ \\{3,6\\}$, which clearly satisfies the requirements. now assume that we have a set $ \\{a_1, ...,a_k\\}$ of distinct positive integers that satisfy the requirement. we wish to enlarge this set by one element into a new set that also works. note that multiplying the entire set by the same positive integer, the conditions still apply: if $ a \\plus{} b|ab$, then $ ra \\plus{} rb|rab|ra\\cdot rb$. let $ r \\equal{} (a_1 \\plus{} 1)(a_2 \\plus{} 1)\\cdots (a_k \\plus{} 1)$. we claim that the set $ \\{ra_1, ra_2,...,ra_k,r\\}$ satisfies the requirements. any pair of $ ra_i$'s satisfies the divisibility requirement by our previous observations. also, for every $ i$, $ r \\plus{} ra_i \\equal{} r(a_i \\plus{} 1)|r^2|r\\cdot ra_i$. it's also clear that the elements of our new set are distinct: all the $ ra_i$'s are distinct b/c the $ a_i$ are distinct, and for every $ i$, $ r\\neq ra_i$ because all $ a_i$ are different from 1. this completes the induction and the proof.", + "Solution_2": "For the first part, I have a different appoach. I hope it may help solving the second part.\r\n\r\nNotice that $ a \\plus{} b | ab$ iff there exists $ n\\in\\mathbb{N}^*$ such that $ ab \\equal{} n(a \\plus{} b)$ or $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\equal{} \\frac {1}{n}$. Now let $ a_i : \\equal{} \\frac {s}{b_i}$ where $ b_i$ are known positive integers and $ s$ will be chosen later. We notice that $ s$ is the same for every $ i$ and $ s$ and $ b_i$ can be not co-prime. We have $ s$ is divisible by every $ b_i$. Moreover, $ \\frac {1}{a_i} \\plus{} \\frac {1}{a_j} \\equal{} \\frac {b_i \\plus{} b_j}{s} \\equal{} \\frac {1}{n}$, this happens iff $ s$ is divisible by every sum $ a_i \\plus{} a_j$ for every $ 1 \\leq i < j \\leq k$. Thus, we choose $ s \\equal{} \\prod_{i \\equal{} 1}^{k} b_i \\prod_{1 \\leq i < j \\leq k} (b_i \\plus{} b_j)$. Now each $ a_i : \\equal{} \\frac {s}{b_i}$ provides the set $ \\{a_1, a_2,...,a_k\\}$ satisfies the condition.\r\n\r\nFor the second part, I did not find the answer. May be there does not exists such set of positive integers $ \\{a_i\\}$. But if there exists, we may construct the set with the same notation above and we must have $ \\frac {1}{a_i} \\plus{} \\frac {1}{a_j} \\plus{} \\frac {1}{a_ia_j} \\equal{} \\frac {1}{n}$, meanwhile $ s^2$ is divisible by every $ s(b_i \\plus{} b_j) \\plus{} b_ib_j$ for every $ 1 \\leq i < j \\leq k$. I did not find out whether the answer is \"Yes\" or \"No\" :(", + "Solution_3": "Cool! My solution is exacts the same as pleurestique, and of course I got stuck with the second part. The thing is, in my and pleurestique's approach, when we are to add the $ k\\plus{}1$-th element to the previously constructed set $ \\{a_1, ...,a_k\\}$, we have two free variables to determine, the new element, lets call it $ x$ and another variable $ y$ which can be multiplied to any $ a_i$ ($ i\\equal{}1,2,\\cdots,k$). The condition $ a_i\\plus{}a_j | a_ia_j$ remains true when both variables are multiplied by the integer $ y$. But in the second part, I added one extra term \"+1\" to the divisibility condition which dismisses the role of $ y$. So we are left with only one variable $ x$ which is not enough to carry on with the induction. On the other hand the approach of Mashimaru seems more structural to handle the second part. As a matter of fact, for a split second I thought I solved the second part by applying Mashimaru's approach twice, but alas, there was a bug in it. I'll try more on this problem. Anyway, the general problem which I was studying was this: Given two integral two-variable polynomials $ p(x,y)$ and $ q(x,y)$ with $ degree(q) > degree(p)$, for any integer $ k$, are there $ k$ distinct integer $ a_1,a_2,\\cdots,a_k$ with $ p(a_i,a_j) | q(a_i,a_j)$ for any $ 1\\leq i\\not\\equal{} j \\leq k$? Another version of this problem ask for an infinite set, formally: when there are infinite distinct integers $ a_1,a_2,\\cdots$ with the aforementioned condition? Of course for the special cases stated in the original post, the infinite version answers No (why?). To make the infinite more practical, one can restrain the condition to only successive integers (a routine kind of restriction in Olympiad problems(, i.e., $ p(a_i,a_{i\\plus{}1}) | q(a_i,a_{i\\plus{}1})$. By inspiration from this general form, I posed the problem posted at: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315281. However in this problem I involved three successive members instead of two and additionally I added another routine restriction of $ a_i$s being a permutation of $ \\mathbb{N}$." +} +{ + "Tag": [ + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "Prove that any set of $ n$ integers has an nonempty subset whose sum is divisible by $ n$.", + "Solution_1": "Consider the numbers:\r\n\r\n$ \\sum_{i=1}^{1}\\(a_i$\r\n$ \\sum_{i=1}^{2}\\(a_i$\r\n$ \\sum_{i=1}^{3}\\(a_i$\r\n\r\n\r\n\r\n$ \\sum_{i=1}^{n}\\(a_i$\r\n\r\nWe have this $ n$ numbers, if there are two of them which are equal $ mod n$ then if we sustract them, we find a nonempty subset whose sum is divisible by $ n$\r\nIf there doesn't exist two that are equal $ mod n$ they are all distinct, and since there are $ n$ numbers, there must be one of them which is divisble by $ n$" +} +{ + "Tag": [ + "function" + ], + "Problem": "Suppose the function $f$ satifies $f(x) + f(x-1) = x^2$ and $f(19) = 94$. What is $ f(94)$? This eludes me for now. Thanks for any help guys.", + "Solution_1": "I think this is right:\r\n\r\nWe have $f(x)+f(x-1)=x^2$\r\n\r\nSo:\r\n\\[f(94)+f(93)=94^2\\]\r\n\r\nWe know that $f(93)+f(92)=93^2\\Rightarrow f(93)=93^2-f(92)$\r\n\r\nContinuing in this manner I believe you get\r\n\r\n\\[f(94)+93^2-92^2+...-20^2+f(19)=94^2\\]\r\n\r\nSo:\r\n\r\n\\[f(94)=94^2-93^2+92^2+...-20^2-94=4181\\]\r\n\r\nI hope I didn't do anything wrong...I m a little tired now.", + "Solution_2": "oops misread the question, should go...\r\nif x is 20 then\r\nf(20)+f(19)=20^2\r\nf(20)+94=400\r\nf(20)=306\r\n\r\nif x is 21 then\r\nf(21)+f(20)=21^2\r\nf(21)+306=441\r\nf(21)=135\r\n\r\nso f(94)=94^2-93^2+92^2....+20^2-94\r\n\r\nand, as my physics teacher would say, its just monkey work from there", + "Solution_3": "Which in turn is just (94+93)+(92+91)+(90+89)+...+20^2+94", + "Solution_4": "Gee... Thanks! :) I can't believe how I didn't see that earlier... :blush: . Hehe. It was on last year's Lehigh Test for Pennsylvania... so I was doing it today.", + "Solution_5": "[quote=\"white_horse_king88\"]Gee... Thanks! :) I can't believe how I didn't see that earlier... :blush: . Hehe. It was on last year's Lehigh Test for Pennsylvania... so I was doing it today.[/quote]\r\n\r\nInteresting. The exact question was number 3 on the 1994 AIME." +} +{ + "Tag": [ + "AMC", + "AIME", + "geometry", + "perimeter", + "rectangle" + ], + "Problem": "ABCD is a rectangle. P, Q, R, S lie on the sides AB, BC, CD, DA respectively. PB = 15, BQ = 20, PR = 30, QS = 40. Find the perimeter of ABCD. \r\n\r\n\r\n\r\nIdeas?", + "Solution_1": "Definitely lacking information, there -- for example, it's easy to make it work with a 30x40 rectange, and you clearly have wiggle room to shrink either of the sides without changing the other. This is rather problematic. Try and find a source other than Kalva.", + "Solution_2": "Urf, yeah, I suspected so, since PR and QS can pretty much be anything if we move R and S around on their respective lines." +} +{ + "Tag": [ + "calculus", + "integration", + "geometry", + "3D geometry", + "algebra", + "factorization", + "sum of cubes" + ], + "Problem": "In terms of [i]n[/i], how many squares have all of its vertices in an [i]n[/i] by [i]n[/i] grid of lattice points?", + "Solution_1": "[hide=\"MAML\"]\nSide-lengths can range from $1$ through $n-1$.\n\nLook first at the squares with edges parallel to the grid lines. For side length $k$, there are $(n-k)^{2}$ such squares.\n\nWithin each such square, we can inscribe $k-1$ distinct tilted squares. Thus, there are $(n-k)^{2}+(k-1)(n-k)^{2}=k(n-k)^{2}$ distinct squares [i]corresponding[/i] to each integral side length.\n\nThe desired sum is $\\sum_{k=1}^{n-1}k(n-k)^{2}$\n\nNote that this is equal to $\\sum_{k=1}^{n}k(n-k)^{2}$, so we can set them equal and perform an index shift to achieve our answer (or we can brute force it out with sum of cubes, sum of squares, sum of integers formulae).\n\nWe get $S(n)=\\sum_{k=1}^{n-1}k(n-k)^{2}=\\frac{n^{4}-n^{2}}{12}$\n[/hide]", + "Solution_2": "What if it was an MxN grid ?\r\n\r\nand wait.. so your formula says that there is one 1 square with vertices consisting of lattice points in a 2X2 grid?\r\n\r\nwoulndt it be ((n+1)^4-(n+1)^2)/12", + "Solution_3": "The Zuton Force is right, that's what I got during the MAML too :lol:", + "Solution_4": "wait.. is a 1 by 1 grid of lattice points a unit square or a point?", + "Solution_5": "It's a point. So the two-by two that you were talking about is really : : <<< a unit square." +} +{ + "Tag": [ + "algebra unsolved", + "algebra" + ], + "Problem": "I can't seem to figure out a nice neat way of going about it...\r\n\r\nIt takes 2 people 130 minutes to clean 3 rooms in the building. how many people will it take to clean the entire 174 rooms in the building if the job must be done in 7.5 hours?\r\n\r\nthere has to be a simple way to solve these... any ideas?", + "Solution_1": "This is the olympiad section; problems of this nature do not belong here. A post of this nature belongs in 'Getting Started'.", + "Solution_2": "2*130/3=260/3\r\n\r\n450x/174=260/3\r\n\r\n1350x=45240\r\n\r\nx=33.5, rounded up to 34\r\n\r\nIn the future, please post these in the Getting Started section." +} +{ + "Tag": [ + "geometry", + "circumcircle", + "rectangle", + "geometry proposed" + ], + "Problem": "Let $ABCD$ be a convex quadrilateral so that there is a point $P$ in it such that $\\angle{PBA}=\\angle{PCD}=90^{0}$ and $\\angle{PAB}+\\angle{PDC}= \\angle{BMC}$ where $M$ is the midpoint of the side $AD$. Prove that $BM=CM$ or $AP=DP$.", + "Solution_1": "I think its not true.\r\nTiks, see if there is anything wrong with this :\r\n\r\nIn the below diagram , $PM \\perp AD$ .\r\nPoints $B$ and $C$ can move on the arcs and satifying the problem conditions, while $BM$ is not necessarily equal to $CM$.", + "Solution_2": "Hey $lomos$,I am sorry I forgot to mention something :oops: .\r\nI edited the post and I hope now it is correct.\r\nThank you for your note.", + "Solution_3": "About the answer of question, how we can construct the configuration of this problem $($ in the case when $AP\\neq DP$ $),$ we restate the notations as follows:\r\n\r\n[b][size=117]PROPOSITION.[/size][/b] \u2013 Let $\\bigtriangleup MBC$ be, an isosceles triangle $($ $MB = MC$ $),$ with it\u2019s circumcircle $(O)$ and also let $P$ be, an arbitrary point inwardly to it. We draw the circle $(O'),$ taken as diameter the segment $PM$ and we denote as $L,$ the intersection point of $(O),$ $(O'),$ the other than $M.$\r\nWe draw two lines through vertices $B,$ $C$ and perpendicular to $PB,$ $PC$ respectively and we denote as $A,$ $D,$ their intersection points, from the segment line $LM.$ Prove that $MA = MD$ and $\\angle PAB+\\angle PDC = \\angle BMC.$\r\n\r\n[b][size=117]PROOF.[/size][/b] - $($ In my drawing $BC = 5.5,$ $MA = MD = 6.1,$ $PB = 2.4,$ $PC = 4.2$ $).$\r\n\r\nWe denote as $M',$ the antidiametric point $($ = antipode $)$ of $M,$ with respect to the circle $(O).$ It is easy to prove that the points $L,$ $P,$ $M',$ are collinear $($ $\\angle MLP = 90^{o}= \\angle MLM'$ $\\Longrightarrow$ $LP\\equiv LM'$ $).$\r\n\r\nFrom cyclic quadrilaterals $BPLA,$ $CPLD,$ $BLMC,$ we have that $\\angle PAB = \\angle PLB$ $,(1)$ and $\\angle PLC = PDC$ $,(2)$ and $\\angle BLC = \\angle BMC$ $,(3)$ and $\\angle PLB = \\angle BMM' = \\angle CMM' = \\angle PLC$ $,(4)$\r\n\r\nFrom $(1),$ $(2),$ $(3),$ $(4)$ $\\Longrightarrow$ $\\angle PAB = \\angle PDC$ $,(5)$ and $\\angle PAB+\\angle PDC = \\angle BMC$ $,(6)$\r\n\r\nWe have now, the configuration of the triangle $PBC,$ with the similar right triangles $\\bigtriangleup BPA$ $($ $\\angle PBA = 90^{o}$ $)$ and $\\bigtriangleup CPD$ $($ $\\angle PCD= 90^{o}$ $),$ erected on it\u2019s sidelines $PB,$ $PC$ respectively, outwardly to it.\r\n\r\nSo, based on the below Lemma, we conclude that the segment line $MM',$ as the midperpendicular of the side segment $BC,$ passes through the midpoint of the segment $AD.$ Hence, we have that $MA = MD$ and the proof is completed.\r\n\r\n[b][size=117]LEMMA.[/size][/b] \u2013 A triangle $ABC,$ is given and let $ABDE,$ $ACFH$ be, two similar rectangles, erected on it\u2019s sidelines $AB,$ $AC$ respectively, outwardly to it. \r\nProve that the midperpendicular of the side segment $BC,$ passes through the midpoint of the segment $DF.$ \r\n\r\nI will post here later the proof I have in mind about this Lemma. \r\n\r\nBecause of $P,$ is the one and only intersection point, of the lines through vertices $B,$ $C,$ and perpendicular to the sidelines $AB,$ $CD$ respectively, in a given quadrilateral $ABCD,$ we can have the configuration as the problem states, if and only if, the right triangles $%Error. \"bigtriangle\" is a bad command.\nBPA,$ $%Error. \"bigtriangle\" is a bad command.\nCPD,$ are similar ones.\r\n\r\nHappy new year to you and all, Kostas Vittas.", + "Solution_4": "Thank you for your solution Vittas! :wink: \r\n\r\nHere is my way.\r\n\r\nContinue the rays $AB$ and $DC$ until intersaction at a point $E$.\r\nIn the triangle $AED$ let $L$ to be the projection of $P$ on the side $AD$ and let $P'$ to be the isogonal point of $P$ in respect to the $AED$.\r\nFrom the problem condition we can easly see that point the quadrilateral $BLMC$ is ciclyc.We also know the well-known fact that the projections of the isogonal points on the sideas of the triangle lie on the same circle( in particular when the isogonal points are the circucentre and orthocentre of the triangle then the mentioned circle is the 9 point circle).So, we have that $M$ is the projection of $P'$ on the side $AD$,therefore $\\angle{P'AD}=\\angle{P'DA}$ =>$\\angle{PAB}=\\angle{P'AD}=\\angle{P'DA}=\\angle{PDC}$=>$\\angle{PAB}=\\angle{PDC}$.\r\n Now if $B'$ and $C'$ are the midpoints of the segments $AP$ and $DP$ respectively,then it is easy to see that $\\triangle{BB'M}=\\triangle{CC'M}$=>$BM=CM$.\r\n\r\nBy the way Vittas,you can see that my last sentence is a prove to your lemma :wink: . \r\n\r\n[color=red][size=150]Happy New Year and Merry Christmas!!!!![/size][/color] :)" +} +{ + "Tag": [ + "inequalities" + ], + "Problem": "Selam alejkum!\r\n\r\nProve that for all positive integers it's true this ineq.:\r\n$(2n^{2}+3n+1)^{n}\\geq 6^{n}(n!)^{2}$", + "Solution_1": "I am not an Iranian but I am still replying for the sake of answering a mathematical problem. I hope that I am not hurting anyone by doing so. If that is not the case then please accept my apologies. \r\n\r\nObservation 1: A.M of the squares of the first $n$ natural numbers is $\\frac{2n^{2}+3n+1}{6}$. \r\n\r\nObservation 2: G.M of the squares of the first $n$ natural numbers is $(n!)^{2/n}$. \r\n\r\nA.M-G.M inequality gives the rest.", + "Solution_2": "You're right! :wink:" +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "$a,b,c>0$. Prove for any $k\\in\\mathbb{R}$\r\n\r\n(1)\t$\\sum(a^2+kbc)^2(a-b)(a-c)\\ge0$. \r\n(2)\t$\\sum(a+k\\frac{bc}a)^2(a-b)(a-c)\\ge0$.", + "Solution_1": "This ineq is based on the idea of Schur ineq. \r\n\r\nLet's see the general version of Schur ineq. \r\n\r\n$A(a-b)(a-c) + B(b-c)(b-a)+C(c-a)(c-b) \\geq 0$ $\\Leftarrow$ $A+C\\geq B$ when $a\\geq b \\geq c$\r\n\r\nThis idea can be easily proved by using $a-c = (a-b)+(b-c)$.\r\n\r\nSo, we can prove the ineqs with proving $A+C \\geq B$. \r\n\r\nLet's start!\r\n\r\nFirst, Let $a\\geq b\\geq c$. \r\n\r\n$A+C \\geq B$ $\\Leftrightarrow$ $(a^2 +kbc)^2 + (c^2 +kab)^2 \\geq (b^2+kca)^2$\r\n\r\n$\\Leftrightarrow 2kabc(a+c-b)+k^2 (b^2c^2 +a^2b^2-c^2a^2)+(a^4+c^4-b^4)\\geq 0$. \r\n\r\nIt's obvious. \r\n\r\nLet's see the second one. \r\n\r\n$A+C\\geq B$ $\\Leftrightarrow$ $(a+ k\\frac{bc}{a})^2 + (c+k\\frac{ab}{c})^2 \\geq (b+k\\frac{ca}{b})^2$\r\n\r\n$\\Leftrightarrow (a^2 +c^2 - b^2)+2k(bc+ab-ca)+k^2a^2 b^2 c^2 (\\frac{1}{a^4} + \\frac{1}{c^4} - \\frac{1}{b^4})\\geq 0$. \r\n\r\nIt's also obvious. \r\n\r\n$Q.E.D.$", + "Solution_2": "[quote=\"Chang Woo-JIn\"]This ineq is based on the idea of Schur ineq. \n\nLet's see the general version of Schur ineq. \n\n$A(a-b)(a-c) + B(b-c)(b-a)+C(c-a)(c-b) \\geq 0$ $\\Leftrightarrow$ $A+C\\geq B$ when $a\\geq b \\geq c$\n\n[/quote]\r\nNice solution.But I think you can't use $\\Leftrightarrow$ here :)", + "Solution_3": "Yep, you are right, :) I corrected.", + "Solution_4": "Sorry I\u2019m late.\r\nIt\u2019s obvious for $k\\ge0$ of course, but what about the case $k<0$ which is far more interesting? ;)", + "Solution_5": "It's not hard.. Just use $AM-GM$. \r\n\r\nbecause we already know $a\\geq b\\geq c$. I thought it is so obvious that I don't need to write it here.. :) ." +} +{ + "Tag": [ + "symmetry" + ], + "Problem": "Right now Im doing AoPS vol 1, and I just finished a chapter on translations/distortions/dilations. Anyways, I was a little confused on some end-of-the chapter problems that required knowledge of point symmetry. I understand line symmetry, but point symmetry just plain confuzzles me. Like why do regular heptagons and equilateral triangles not have point symmetry. Can someone give me a brief explanation of point symmetry?\r\n\r\nThanks! :lol:", + "Solution_1": "Ooh, I just finished it too (Well, I'm on the end of chapter problems, but who cares)!\r\nIf I'm not completely wrong:\r\nWell, by points symmetry, basically it connects one vertex to another, so there will be always be one left in polygons with an odd number of sides. (Of course, the leftover vertex could be the point of symmetry, but then that messes up the polygon)", + "Solution_2": "[url]http://www.regentsprep.org/Regents/math/symmetry/Psymmet.htm[/url]", + "Solution_3": "thanks! that makes more sense!" +} +{ + "Tag": [ + "function", + "vector", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Let $n$ be at least $2$ and $f$ be a $C^1$ function from $R^n$ to $R$ such that the set $A$ of its zeros is nonempty, bounded and in each point $x$ of $A$, $df(x)$ is non zero. Prove that for any vector $x$ of norm $1$ one can find $y$ in $A$ such that $x=\\frac{gradf(y)}{||gradf(y)||}$.", + "Solution_1": "I think for example that for the case $n=2$, it is quite intuitive:\r\n\r\nThe set $A$ is bounded so one can suppose that $f(x)>0$ for $|x| \\geq R$.\r\nBecause $A$ is non empty and $grad(f) \\neq 0$ for any $x \\in A$, this shows that there exists at least one $x \\in \\mathbb{R}^2$ such that $f(x)<0$. WLOG, $f(0)<0$.\r\n\r\nNow, for any $a \\in [0; 2 \\pi[$, $g_a(t)=f(t cos(a); t sin(a))$ is continuous on $[0;R]$, $g_a(0)<0$, $g_a(R)>0$. \r\nCall $Z_a=(t_a cos(a); t_a sin(a))$ the nearest zero of $f$ on this line. (i.e. $t_a=inf_{[0;R]} \\{t | g_a(t)=0\\}$).\r\nHence $Z: [0; 2 \\pi[ \\to \\mathbb{R}^2$, $Z(a)=Z_a$ is a continuous closed curve that turns around $0$.\r\nFrom here, it is quite straightforward (need a proof but it's clear on a picture) that for any unit vector $x$, there exists some $a \\in [0;2 \\pi[$ such that $x=\\frac{gradf(Z(a))}{||gradf(Z(a))||}$.\r\n\r\nSo in fact this problem is equivalent to the following one: \r\nin $\\mathbb{R}^n$, suppose that $S$ is a surface diffeomorphic to $\\mathbb{S}^{n-1}$ and $f: \\mathbb{R}^n \\to \\mathbb{R}$ is a $C^1$ function such that $grad(f)$ never vanishes on $S$. Shows that for any unit vector $x$, there exists some $y \\in S$ such that $x=\\frac{gradf(y)}{||gradf(y)||}$.\r\n\r\n(I think it is easy because $S$ is diffeomorphic to $\\mathbb{S}^{n-1}$, and it is true if $S=\\mathbb{S}^{n-1}$ )", + "Solution_2": "[quote=\"alekk\"]I think for example that for the case $n=2$, it is quite intuitive:[/quote]As intuitive as Jordan's curve theorem :)", + "Solution_3": ":D but here everything is \"regular\", so proofs are easier ..\r\n(even Jordan theorem is quite easy for regular curves :) )", + "Solution_4": "[quote=\"harazi\"]Let $n$ be at least $2$ and $f$ be a $C^1$ function from $R^n$ to $R$ such that the set $A$ of its zeros is nonempty, bounded and in each point $x$ of $A$, $df(x)$ is non zero. Prove that for any vector $x$ of norm $1$ one can find $y$ in $A$ such that $x=\\frac{\\nabla f(y)}{||\\nabla f(y)||}$.[/quote]\r\nGiven $x \\in S^{n-1}$, minimize / maximize $g(y) = y^Tx$ on the set $A$ and use the Lagrange multiplier theorem. That will give us two points in $A$ on which $x=\\pm \\frac{\\nabla f(y)}{||\\nabla f(y)||}$. Then it's only a matter of finding the right sign :P" +} +{ + "Tag": [ + "inequalities" + ], + "Problem": "I am trying to follow this lecture (http://www.tjhsst.edu/~tmildorf/math/Inequalities.pdf) on inequalities, but I am very confused with the asterisks everywhere..\r\n\r\nI dont know if they mean multiplication or what as the asterisk is used in many places lines like:\r\n\r\n$(a+b)^{2}= k^{2}= (a * b)^{2}+4ab$\r\n\r\nthe asterisk in this case should be a subtract symbol or in cases where ... shoudl be used *** is used\r\n\r\n$a_{1}, a_{2}, ***, a_{n}$\r\n\r\nam I just really dumb, or is there actually an error? thanks", + "Solution_1": "What are you using to view the pdf file? It seems to be working fine without those asterisks here.", + "Solution_2": "I am using adobe reader.. :(", + "Solution_3": "I don't see any asterisks...", + "Solution_4": "wow this is really lame.. it seems to be only my computer that is doing this, I am using adobe reader 8.0.0 to view PDF files.\r\n\r\nsorry about this, time to find an answer :ninja:", + "Solution_5": "here's what I see:\r\n[img]7967[/img]\r\n\r\nedit: judging by the thick black lines, you might need a post script reader like gsview... it's free", + "Solution_6": "yeah for me I see $***$ instead of $..., >,-$\r\n\r\nits really annoying.", + "Solution_7": "Don't use Acrobat Reader.\r\nUse FoxitReader." +} +{ + "Tag": [], + "Problem": "uspho selection process\r\nwhat is the first and second tests on(before semifinals)? both mechanics and electromagnetism?", + "Solution_1": "The first, only on mechanics. The second, on both.\r\n\r\nI think Google could have answered that question for you.", + "Solution_2": "I do not believe that is correct. I think the first test is a multiple-choice mechanics test, the second test is a free-response mechanics test, and the third test (the semifinal) covers most of the olympiad syllabus.", + "Solution_3": "Yeah, the first one is a multiple-choice mechanics test. The second one is an hour-long free-response test that covers mostly mechanics, but it does have some electrostatics (this year, it had one electrostatics question and three on mechanics)." +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "Let a,b,c be lengths of a triangle.\r\nProve that\r\n$ R \\ge \\frac{{\\sqrt {a^2 (p \\minus{} b)(p \\minus{} c) \\plus{} b^2 (p \\minus{} a)(p \\minus{} c) \\plus{} c^2 (p \\minus{} a)(p \\minus{} b)} }}{p}$ :lol:", + "Solution_1": "[quote=\"TRAN THAI HUNG\"]Let a,b,c be lengths of a triangle.\nProve that\n$ R \\ge \\frac {{\\sqrt {a^2 (p \\minus{} b)(p \\minus{} c) \\plus{} b^2 (p \\minus{} a)(p \\minus{} c) \\plus{} c^2 (p \\minus{} a)(p \\minus{} b)} }}{p}$ :lol:[/quote]\r\n\r\n\r\nsubstitution $ p \\equal{} x \\plus{} y \\plus{} z,a \\equal{} x \\plus{} y,b \\equal{} y \\plus{} z,c \\equal{} z \\plus{} x$\r\n\r\nThe following inequality is equivalent to \r\n$ \\frac{1}{16}\\,{\\frac {\\left( x \\plus{} y \\right) ^{2} \\left( y \\plus{} z \\right) ^{2} \\left( z \\plus{} x \\right) ^{2}}{\\left( x \\plus{} y \\plus{} z \\right) zxy}} \\minus{} {\\frac {\\left( x \\plus{} y \\right) ^{2}xy \\plus{} \\left( y \\plus{} z \\right) ^{2}yz \\plus{} \\left( z \\plus{} x \\right) ^{2}zx} {\\left( x \\plus{} y \\plus{} z \\right) ^{2}}}\\geq 0$\r\n\r\n$ \\Longleftrightarrow (x \\plus{} y \\plus{} z)(z^2y \\plus{} y^2z \\plus{} z^2x \\minus{} 6zxy \\plus{} y^2x \\plus{} x^2z \\plus{} x^2y)^2\\geq 0$\r\n\r\nobvious true", + "Solution_2": "[quote=\"TRAN THAI HUNG\"]Let a,b,c be lengths of a triangle.\nProve that\n$ R \\ge \\frac {{\\sqrt {a^2 (p \\minus{} b)(p \\minus{} c) \\plus{} b^2 (p \\minus{} a)(p \\minus{} c) \\plus{} c^2 (p \\minus{} a)(p \\minus{} b)} }}{p}$ :lol:[/quote]\r\n\r\n\\[ \\Longleftrightarrow p^2 R^2 \\ge p^2 \\sum\\limits_{cyc} {a^2 } \\minus{} p\\sum\\limits_{sym} {a^2 b} \\plus{} abc\\sum\\limits_{cyc} a\\]\r\n\\[ \\Longleftrightarrow p^2 R^2 \\ge p^2 \\left( {2p^2 \\minus{} 8Rr \\minus{} 2r^2 } \\right) \\plus{} p\\left( {12pRr \\minus{} 2p\\left( {p^2 \\plus{} 4Rr \\plus{} r^2 } \\right)} \\right) \\plus{} 8p^2 Rr\\]\r\n\\[ \\Longleftrightarrow \\left( {R \\minus{} 2r} \\right)^2 \\ge 0.\\]\r\n\r\nclear..." +} +{ + "Tag": [], + "Problem": "Can anyone tell me what NEAML or MAML is and how to join? Or does your school have to participate for you to?", + "Solution_1": "[quote=\"alkjash\"] does your school have to participate for you to?[/quote]\r\n\r\n[color=black]Yes.[/color]", + "Solution_2": "MAML and NEAML are the Massachusetts and New England high school math leagues/meets. When you get to ABRHS and join the math team, they'll compete in the local Massachusetts Math League (MML), and in March and April, we'll go to MAML and then NEAML. So basically it's part of the ABRHS math team.", + "Solution_3": "[quote=\"calc rulz\"]in March and April, we'll go to MAML and then NEAML.[/quote]\r\n\r\nCocky much?", + "Solution_4": "Ok, thanks. So it's in High School and I don't have to worry much. Great", + "Solution_5": "Ya, it's high school.\r\nI know that in GBML, some number of top schools (at least 3) qualify for MAML.\r\nIn SMML, some top schools also qualify for MAML ( not sure how many.) Also, #1 school from that league anomatically goes to NEAML and I don't know if that happens in GBML.\r\nGBML and SMML are math leagues, also high school ones." +} +{ + "Tag": [ + "Functional Analysis", + "advanced fields", + "advanced fields unsolved" + ], + "Problem": "In a complex separable Hilbert space with orthonormal basis $e_{0},e_{1},\\ldots$, let $U$ (the [i]unilateral shift[/i]) be the operator defined by $Ue_{n}=e_{n+1},\\ \\forall n\\ge 0$.\r\n\r\nIs $U$ a finite product of normal operators?", + "Solution_1": "No, it isn't. Indeed, if $N$ is normal and $\\|Nx\\|\\ge c\\|x\\|$ for all $x$ with some $c>0$, then $N$ must be invertible. It follows that every operator in the product must be invertible. But then the whole product must be invertible, but it isn't." +} +{ + "Tag": [ + "geometry", + "geometry proposed" + ], + "Problem": "[color=darkblue] Let $ ABC$ be a triangle. Find all segments $ XY$, which are situated in the interior of the triangle, such that $ XY$ and five from the segments $ XA,XB,XC$, $ YA,$ $ YB$, $ YC$ divide area of $ ABC$ in five equivalent parts. Prove that all found segments are concurrent.\n[/color]", + "Solution_1": "$[\\triangle AXB] \\equal{} [\\triangle CXB] \\equal{} [ \\triangle AYX] \\equal{} [\\triangle CYX] \\equal{} [\\triangle CYA]$ $\\Longrightarrow$ $ X$ lies on the B-median $BM_b$ and $ Y$ coincides with the centroid of $ \\triangle AXC.$ Further, $B,X,Y$ are collinear such that $ BX \\equal{} \\frac {_2}{^5}BM_b$ and $ YM_b \\equal{} \\frac {_1}{^5}BM_b.$ By similar reasoning, we conclude that such three segments concur at the centroid $G$ of $\\triangle ABC.$" +} +{ + "Tag": [ + "function", + "logarithms", + "limit", + "calculus", + "calculus computations" + ], + "Problem": "show the following 2 identities... :| \r\n\r\n\r\n$ \\textbf{(i)}\\;\\;\\;\\sum_{n=2}^\\infty\\;\\frac{(-1)^{n}\\;\\zeta{(n)}}{n}\\;=\\;\\boxed{\\gamma}$\r\n\r\n\r\n$ \\textbf{(ii)}\\;\\;\\;\\sum_{n=2}^\\infty\\;\\frac{\\left( \\zeta{(n)}-1\\right) }{n}\\;=\\;\\boxed{1-\\gamma}$\r\n\r\n\r\nwhere $ \\gamma$ is [b]Euler's[/b] constant.", + "Solution_1": "1. By alternating series test, this series clearly converges. Now let's consider the function $ f(x) = \\sum_{n=2}^{\\infty}\\frac{(-1)^{n}\\zeta(n)}{n}x^{n}$. For $ |x|<1$, it absolutely converges. So\r\n\r\n\\begin{eqnarray*}f(x) & = & \\sum_{n=2}^{\\infty}\\frac{(-1)^{n}\\zeta(n)}{n}x^{n}\\\\ & = & \\sum_{n=2}^{\\infty}\\sum_{m=1}^{\\infty}\\frac{(-1)^{n}}{n}\\left( \\frac{x}{m}\\right)^{n}\\\\ & = & \\sum_{m=1}^{\\infty}\\sum_{n=2}^{\\infty}\\frac{(-1)^{n}}{n}\\left( \\frac{x}{m}\\right)^{n}\\\\ & = & \\sum_{m=1}^{\\infty}\\left[ \\frac{x}{m}-\\ln\\left( 1+\\frac{x}{m}\\right) \\right]. \\end{eqnarray*}\r\n\r\nSince $ \\gamma = \\sum_{m=1}^{\\infty}\\left[ \\frac{1}{m}-\\ln\\left( 1+\\frac{1}{m}\\right) \\right]$ and $ |\\ln(1+x)-x| \\leq Mx^{2}$ for suffciently small $ x$ and for some $ M \\geq 0$, \r\n\r\n\\begin{eqnarray*}| \\gamma-f(x) | & = & \\left| \\sum_{m=1}^{\\infty}\\left[ \\frac{1-x}{m}+\\ln\\left( 1-\\frac{1-x}{m+1}\\right) \\right] \\right|\\\\ & \\leq & \\sum_{m=1}^{\\infty}\\left| \\frac{1-x}{m}+\\ln\\left( 1-\\frac{1-x}{m+1}\\right) \\right|\\\\ & = & \\sum_{m=1}^{\\infty}\\left| \\frac{1-x}{m(m+1)}+\\frac{1-x}{m+1}+\\ln\\left( 1-\\frac{1-x}{m+1}\\right) \\right|\\\\ & \\leq & \\sum_{m=1}^{\\infty}\\left[ \\frac{1-x}{m(m+1)}+\\left| \\frac{1-x}{m+1}+\\ln\\left( 1-\\frac{1-x}{m+1}\\right) \\right| \\right]\\\\ & \\leq & \\sum_{m=1}^{\\infty}\\left[ \\frac{1-x}{m(m+1)}+\\frac{M(1-x)^{2}}{(m+1)^{2}}\\right]\\\\ & = & (1-x)\\left( 1+M\\frac{\\pi^{2}-6}{6}(1-x) \\right) \\end{eqnarray*}\r\n\r\nif $ x$ approaches 1 from the left side.\r\n\r\nTherefore, $ \\gamma = \\lim_{x\\to1^{-}}f(x) = \\sum_{n=2}^{\\infty}\\frac{(-1)^{n}\\zeta(n)}{n}$ by Abel's theorem." +} +{ + "Tag": [ + "LaTeX", + "floor function" + ], + "Problem": "This is a kind of embarrassing question, but I can't for the life of me figure out how to get brackets to show up correctly in a LaTeX document. I've tried \\brace, but I can't get anything to show up inside it without looking rather strange. Could someone help me out?", + "Solution_1": "\\{ and \\} should do it.", + "Solution_2": "Wow, I can't believe I didn't think of that. Thanks for the help.", + "Solution_3": "Is there anyway we can enlarge the curly bracket ${(\\sum_{i=1}^{n}A_{i})}$ to make it looks nice and a bit \"square\" in the middle? :oops:", + "Solution_4": "Do you mean like this?\r\n$\\left\\{\\sum_{i=1}^{n}A_{i}\\right\\}$\r\nAny bracket can be enlarged using \\left and \\right. In this case you need to replace \\{...\\} with \\left\\{...\\right\\}", + "Solution_5": "I want the $(\\ldots )$ to be enlarged and looks like a square with the middle looks straight and the edges looks like being cut into curly.", + "Solution_6": "Can you post a link to an example of this? Is it in [url=http://tug.ctan.org/tex-archive/info/symbols/comprehensive/]The Comprehensive LaTeX Symbol List[/url]?", + "Solution_7": "[quote=\"bchui\"]I want the $(\\ldots )$ to be enlarged and looks like a square with the middle looks straight and the edges looks like being cut into curly.[/quote]\r\n\r\nYou mean normal brackets? $\\left [ \\sum_{n=bleh}^{\\pi}\\left\\lfloor \\pi\\right\\rfloor\\right ]$", + "Solution_8": "solafidefarms: The forum made a mess of your posting. If you leave out the \\ before the [ and the ] you get\r\n$\\left[\\sum_{n=bleh}^{\\pi}\\left\\lfloor \\pi\\right\\rfloor\\right]$\r\nIs that what you meant?", + "Solution_9": "Yes, it is; argh but that's ugly.", + "Solution_10": "It is the symbol pair ${ \\lgroup \\sum_{i=1}^{n}A_{i}\\rgroup }$ I was looking for, but I still don;t know how to vary the size of the usual bracket $(\\ldots )$ to make it larger.\r\n\r\nHold on, how come the upper half of it does not appear?", + "Solution_11": "[quote=\"bchui\"]I still don;t know how to vary the size of the usual bracket $(\\ldots )$ to make it larger.?[/quote]But you do because I told you how to above where I said:\n[i]Any bracket can be enlarged using \\left and \\right[/i]\n[quote=\"bchui\"]Hold on, how come the upper half of it does not appear?[/quote]Again, I asked you to look at The Comprehensive LaTeX Symbol List where it says:\r\n[i]These symbols must be used with \\left and \\right.[/i]\r\nso again telling you what to do.\r\n\r\nThus you get:\r\n$\\left\\lgroup \\sum_{i=1}^{n}A_{i}\\right\\rgroup$" +} +{ + "Tag": [ + "inequalities", + "inequalities unsolved" + ], + "Problem": "If $ x,y,z>0$ prove that \\[ \\sum \\frac{x}{x^2 \\plus{}3} \\leq \\frac{3}{4},\\] where the sum is cyclic.", + "Solution_1": "I think we should have the condition $ xyz\\equal{}1$.You can see [hide=\"here\"]http://www.mathlinks.ro/viewtopic.php?t=39571[/hide]", + "Solution_2": "[quote=\"matha\"]If $ x,y,z > 0$ prove that\n\\[ \\sum \\frac {x}{x^2 \\plus{} 3} \\leq \\frac {3}{4},\\]\nwhere the sum is cyclic.[/quote]\r\nThis is only true for $ x \\plus{} y \\plus{} z\\leq 3$. If $ x \\plus{} y \\plus{} z\\geq 3;$ set $ x \\equal{} 2 \\equal{} y; z \\equal{} 1$. Then LHS=$ \\frac 47 \\plus{} \\frac 14 > \\frac 34$ ;) \r\nSo I prove it for $ x \\plus{} y \\plus{} z\\leq 3$.\r\nRewrite the given inequality as\r\n\\[ \\sum_{cyc}\\frac x {2x \\plus{} 2}\\leq \\frac 34 \\iff \\sum_{cyc}\\frac x{x \\plus{} 1}\\leq \\frac 32\\]\r\n\r\n\\[ \\iff \\sum_{cyc}\\frac 1{x \\plus{} 1}\\geq \\frac 32\\]\r\nWhich is perfectly true, since from Cauchy we have\r\n\\[ \\sum_{cyc}\\frac 1{x \\plus{} 1}\\geq \\frac 9{x \\plus{} y \\plus{} z \\plus{} 3}\\geq \\frac 32[\\because x \\plus{} y \\plus{} z\\leq 3]\\]\r\nEquality holds for $ x\\equal{}y\\equal{}z\\equal{}1$.\r\nQED. :)\r\n[edit: Sorry superman , I did not see your post :oops: ]", + "Solution_3": "Hey Potla,\r\nI like your phrase \"perfectly true\" haha.", + "Solution_4": "[quote=\"new_member\"]Hey Potla,\nI like your phrase \"perfectly true\" haha.[/quote]\r\nThis might be off-topic; but I think I use that phrase too often.... :rotfl: :lol:" +} +{ + "Tag": [ + "linear algebra", + "matrix", + "linear algebra unsolved" + ], + "Problem": "Prove that if $A,B,C$ be matrices in $M_{n,p}(K); M_{p,q}(K); M_{q,r}(K)$ then \r\n$rank(ABC) \\leq rank(B)$", + "Solution_1": "$rank(AB)\\leq \\min\\{rank(A),rank(B)\\}$." +} +{ + "Tag": [ + "combinatorics proposed", + "combinatorics" + ], + "Problem": "Let $A_1A_2A_3A_4A_5A_6A_7A_8$ be convex 8-gon (no three diagonals concruent).\r\nThe intersection of arbitrary two diagonals will be called \"button\".Consider the convex quadrilaterals formed by four vertices of $A_1A_2A_3A_4A_5A_6A_7A_8$ and such convex quadrilaterals will be called \"sub quadrilaterals\".Find the smallest $n$ satisfying:\r\nWe can color n \"button\" such that for all $i,k \\in\\{1,2,3,4,5,6,7,8\\},i\\neq k,s(i,k)$ are the same where $s(i,k)$ denote the number of the \"sub quadrilaterals\" has $A_i,A_k$ be the vertices and the intersection of two its diagonals is \"button\".", + "Solution_1": "I think it's hard. I guess n=14", + "Solution_2": "In my exercise , I showed for n =16 .", + "Solution_3": "Let $n$ be minimal such that all conditions are satisfied. A button covers ${4\\choose 2}$ pairs of vertices, so\r\n\r\n${4\\choose 2} n = \\sum_{i0 then if 7/3^n-1 and 7/3^n+1 then 7/2 which is false! Then we must have 7^n/3^n+1 or 7^n/3^n-1. This also a contradiction, since 3^n+1<7^n for n>=1. Then n=0 is the only soln.", + "Solution_5": "Regarding the first one: \r\nDenote 2^n by a. Then a/3^n-1. Since (3, a)=1 we have a/3^(fi(a))-1, where fi(a) is Euler's indicator. Since a_k= 3^k-1 is a Mersenne sequence we must have a/3^(n, fi(a))-1. Write n=2^p*t where t is an odd number. Then, because fi(a)=2^(n-1), we have (n, fi(a))=2^p. So we got 2^n/ 3^(2^p)-1, so 2^(2^p*t)/3^(2^p)-1. This means that 2^(2^p)/3^(2^p)-1. Let's see what happens for small values of p. p can be 0 and we get 2/4, for p=1 we have 4/8 and for p=2 we get 16/80. These are all true. We can show that for p>2 this is impossible. If 2^(2^p)/3^(2^p)-1 and 2^(2^(p+1)) does not divide 3^(2^p)-1 or 2^(2^p) does not divide 3^(2^p)-1 we have 2^(2^(p+1)) does not divide 3^(2^(p+1))-1. Check it out and you'll see why; use the fact that 3^(2^(p+1))-1=(3^(2^p)-1)*(3^(2^p)+1) and the fact that 2/3^(2^p)+1 but 4 does not divide it. We can also see that for p= 0, 1 or 2 and t odd, t>=3 we have no soln. So the only solns are n=0(the trivial case), n=1, n=2, n=4. \r\n\r\nHope it's right and I made myself understood. See ya! :)", + "Solution_6": "Thanks a lot !\r\nI didn't expect the second one to be so easy ... \r\nYou know, I didn't make lots of real progress for the first one so I didn't spend much time to the second one, because the second one looked more difficult than the first one.", + "Solution_7": "I think I have a bit different solution for the first one:\r\n\r\nAt first we look for odd n. Then we get one solution n = 1. (Arne already showed how we get it)\r\nSo we are looking now for an even n. Let n be n = u*2^k, where k >= 1 and u is odd. Then we can prove (by induction for example) that when n = u*2^k then 3^n - 1 is divisible by 2^(k + 2) but not by 2^(k + 3). But \r\nwe are looking for n so that 2^n = 2^(u*2^k) divides 3^n - 1.\r\nSo we get k + 2 >= u*2^k, easily we can show now that u must be 1 (when u > 1, then u*2^k is always > k + 2). Now we have to find\r\nsolutions for k + 2 >= 2^k, only positive integers that fit are k = 1 (3 > 2) and k = 2 (4 = 4). And so we get the solutions n = 1*2^1 = 2 and n = 1*2^2 = 4. So we have managed to find all solutions in positive integers (n = 1, n = 2, n = 4)\r\n\r\nhowever I am not sure if this solution is correct:)" +} +{ + "Tag": [ + "blogs", + "search", + "\\/closed" + ], + "Problem": "Are you allowed to have two blogs and post on both?", + "Solution_1": "I don't think so.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=820457910&t=224298", + "Solution_2": "Okay, thanks. Would it be bad if I made a new account and made a blog with that account?", + "Solution_3": "We will probably allow multiple blogs once AoPS 2.0 is out." +} +{ + "Tag": [ + "function", + "integration", + "logarithms", + "inequalities", + "real analysis", + "real analysis solved" + ], + "Problem": "Suppose that the function $f(x)$ is monotone increasing and continuous and $f(0)=0$.\r\nLet $a>0,f(a)\\geqq b>0$,prove that \r\n\r\n\\[ab\\leqq\\int^a_0 f(t)dt+\\int^b_0 f^{-1}(t)dt\\]\r\n\r\nThen for $x>1,y>0$ ,prove that \r\n\r\n\\[xy\\leqq x( \\log x-1)+e^y\\]", + "Solution_1": "Young's Inequality has a nice well known geometrical proof.\r\nAnd second inequality is just an application for $f(x)=\\ln(x+1)$.\r\n\r\nVery nice approach, kunny!\r\nBut how could we see it in your initial problem ;) :P" +} +{ + "Tag": [ + "AMC" + ], + "Problem": "The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is \r\n\r\n$ \\textbf{(A)}\\ 11 \\qquad \\textbf{(B)}\\ 12 \\qquad \\textbf{(C)}\\ 13 \\qquad \\textbf{(D)}\\ 14 \\qquad \\textbf{(E)}\\ 15$", + "Solution_1": "we know there has to be 3 8s, 2 of the lowest number we can get, one more than that, and then, that number plus 8. the last one will be used to make the mean equal 8. 4 is too low for the lopwest number so we try 5. the highest needs to be 13. we have 5 5 6 8 8 8 _ _ 13. the 2 left blank can be used to make the mean equal 8. so 13", + "Solution_2": "[quote=\"iin77\"]The answer's 14.\n\n6,6,6,8,8,8,8,14[/quote]\r\n\r\nPlease explain WHY your answer is the correct answer when posting.\r\n\r\n[hide=\"solution\"]If the mean is 8, then the sum of the integers is $ 64$. In addition, because the range is 8, try letting the first number be $ 7$, $ 6$, etc to find the first value that will work.\n\nIf the smallest number is 7, then the largest number is 15. At least two of the numbers must be 8, and of the remaining numbers must be $ \\geq7$, so their sum must be at least $ 7\\plus{}15\\plus{}8\\cdot2\\plus{}7\\cdot4\\equal{}66$.\n\nA sample sequence with smallest number $ 6$ is 6,6,6,8,8,8,8,14, so 14 is largest integer that can be in the set.[/hide]", + "Solution_3": "what is the difference between mode and \"unique\" mode", + "Solution_4": "Unique mode merely means that there is only 1 mode.", + "Solution_5": "[quote=\"Hustla25\"]what is the difference between mode and \"unique\" mode[/quote]\r\n\r\nIt is possible that there are two or more numbers that occur the most times (i.e. such as in the set 1,2 or in the set 1,2,2,3,3,4). If that is the case, then the average of the modes is the mode. However, a unique mod guarantees that there is a single number that is the most common.", + "Solution_6": "Wait... The question says [b]integers[/b] so why can't we use negative numbers and make a huge largest number? Perhaps the question meant [b]positive[/b] integers?", + "Solution_7": "That's why they specify the range is 8. So, the largest number occurs when the smallest number is at its maximum.", + "Solution_8": "On the official solution book, they used process of elimination by proving that the answer [i]couldn't[/i] be the largest answer choice, $E. 15 and then showing that it [i]could[/i] be the next largest answer choice, $D. 14$.\n\nHow would you solve this if you wren't given answer choices?", + "Solution_9": "[quote=flyingdragon]On the official solution book, they used process of elimination by proving that the answer [i]couldn't[/i] be the largest answer choice, $E. 15 and then showing that it [i]could[/i] be the next largest answer choice, $D. 14$.\n\nHow would you solve this if you wren't given answer choices?[/quote]\n\nI actually have the same question. All I did in order to find the answer was show that 15 wouldn't work and show that 14 would. I was wondering if anyone knew how to do it without answer choices...", + "Solution_10": "[quote=flyingdragon]On the official solution book, they used process of elimination by proving that the answer [i]couldn't[/i] be the largest answer choice, $E.$ 15 and then showing that it [i]could[/i] be the next largest answer choice, $D. 14$.\n\nHow would you solve this if you weren't given answer choices?[/quote]\n\nI'm not sure. Also fixed your LaTeX\n", + "Solution_11": "[quote=vsar0406][quote=flyingdragon]On the official solution book, they used process of elimination by proving that the answer [i]couldn't[/i] be the largest answer choice, $E. 15$ and then showing that it [i]could[/i] be the next largest answer choice, $D. 14$.\n\nHow would you solve this if you wren't given answer choices?[/quote]\n\nI actually have the same question. All I did in order to find the answer was show that 15 wouldn't work and show that 14 would. I was wondering if anyone knew how to do it without answer choices...[/quote]\n\nLet the integers be $a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8$ with $a_1\\leq\\cdots\\leq a_8$. Using the median$^1$ and range$^2$ information, we can rewrite this list as $$a_8-8,a_2,a_3,8,8,a_6,a_7,a_8.$$ Since $a_8-8=a_1\\leq a_4=8$, we have $a_8\\leq 16$. If $a_8=16$, then $a_1=a_8-8=16-8=8$, and so all the other numbers are at least $a_1=8$. But this is impossible, because $a_8$ is greater than $8$, and so the the mean would be greater than $8$. The next smaller integer is $15$, and we are now in the same place as if we had the answer choices.\n\n[size=75]$^1$ We must have $a_4\\leq a_5$ and $a_5=16-a_4$. If $a_4$ isn't $8$, then the inequalities lock $8$ out of appearing in the list, contradicting it being the mode.\n$^2$ By definition of range, $a_8-a_1=8$.[/size]", + "Solution_12": "[b]Solution:[/b] Clearly both of the middle numbers have to be $8$. So we have something like: $x, y, z, 8, 8, a, b, c$. We know that $c-x=8$, while $a \\le b \\le c$ and $x \\le y \\le z$. Furthermore, we have that $x+y+z+a+b+c = 2c - 8 + y+z+a+b = 48$. Clearly we want to minimize $y+z+a+b$ so we can let $a,b=(8,8)$ and $(y, z) = (x, x)$. So this gives us $2c - 8 + 2x + 16=48 \\Rightarrow 2c + 2x = 40 \\Rightarrow c+x=20$ when $c$ is maximized. Since $c-x=8$, we can deduce that $c=\\boxed{14}$." +} +{ + "Tag": [ + "conics", + "parabola" + ], + "Problem": "Find the equation of the parabola with directrix at x=3 and vertex (0,0).", + "Solution_1": "Parabolas can be written in the form $ x\\minus{}h \\equal{} (\\frac {1}{4c})(y \\minus{} k)^2$ where (h,k) is the vertex and $ c$ is the distance from the vertex to the directriz or focus.\r\n\r\n$ x \\minus{} (0) \\equal{} \\frac {1}{4( \\minus{} 3)}(y \\minus{} (0))^2$ Note that $ c \\equal{} \\minus{} 3$ because the parabola opens to the left.\r\n\r\nSo the equation is: $ x \\equal{} \\frac {y^2}{ \\minus{} 12}$" +} +{ + "Tag": [ + "geometry", + "videos", + "AMC" + ], + "Problem": "Message from George Csicsery, independent documentary film maker and director of the forthcoming film about the 2006 IMO:\r\n\r\n\"Finally, the HARD PROBLEMS trailer has been up on YouTube for several days, and apparently there are some comments already and a decent rating.\r\n\r\nhttp://www.youtube.com/watch?v=d9ZDgjzNeBU\r\n\r\nGeorge\"\r\n\r\nYou might like to take a look at it. I am also posting this in the IMO Forum.\r\n\r\n Steve Dunbar\r\nMAA Director, American Mathematics Competitions", + "Solution_1": "I'd like to add that the world premiere of Hard Problems will be at the 2008 Joint Mathematics Meetings in San Diego on Tuesday, January 8 at 6:00 p.m.\r\n\r\nStudents in high school (or younger) who happen to be in the area can get in for $5 (if you register in advance by December 14) or $10 at the door. This registration fee allows you to attend the entire 4-day Meetings. Although the meetings are largely intended for professional mathematicians, there are some events for students, including the always popular \"Who Wants to be a Mathematician\" game show. \r\n\r\nAs a (slight) added bonus, you can meet some the Art of Problem Solving staff: we'll be at the meetings and will have a booth as part of the Exhibits Hall.\r\n\r\nSee http://www.ams.org/amsmtgs/2109_intro.html for more details.", + "Solution_2": "Thanks for the reminder about the premiere in San Diego. I'm so focused on helping George with the details, it is hard to think that far ahead! Immediately after the premiere, the DVDs of Hard Problems will be for sale at the MAA booth. We'll also be handling some sales later through the AMC office, and the MAA Bookstore will have some too.\r\n\r\nThe premiere should be a lot of fun. The MAA is now doing a premiere of a math related movie at each Joint Math Meeting in January. Last January it was \"Flatland\" which was very well received. There are a lot of interesting books and things at the exhibit hall each year, among which this year will be AoPS.\r\n\r\n-- Steve Dunbar\r\nMAA Director, American Mathematics Competitions", + "Solution_3": "Will we be able to order the DVD's in advance?", + "Solution_4": "No, pricing is not yet completely determined. Available form January onward.\r\n\r\n-- Steve Dunbar\r\nMAA Director, American Mathematics Competitions", + "Solution_5": "i want to go so badly. i am going to convince my parents to take me to the premier. im going even if i have to run all the way from L.A. to San Diego. :D\r\ni love these kinds of documentaries, and the trailer looks great. they always remind me of how passionate some people are, and how they enjoy doing something other than video games or watching senseless t.v. shows. like, i loved SpellBound. \r\n\r\nCAN\"T WAIT!!!!!", + "Solution_6": "Will this \"Hard Problems\" be shown in theater next year?", + "Solution_7": "that's what i want to know because i live in pennsylvania, and it's a pretty long way to san diego...\r\nif you know what i mean... :lol:", + "Solution_8": "This looks awesome. Too bad I won't exactly be able to make it from Chicago. [I will be in San Diego like Nov. 2,3 maybe, but that doesn't help :P]. It is nice for the IMO kids to get some exposure.", + "Solution_9": "I would definitely like to go, but being 6+ hours away and all...\r\n\r\n*Thinks of a random reason that'll convince my parents to travel 400 miles in the middle of January...*", + "Solution_10": "[quote=\"i_like_pie\"]\n*Thinks of a random reason that'll convince my parents to travel 400 miles in the middle of January...*[/quote]\r\n\r\n-It will get you a scholarship to any University you desire\r\n-It is for the good of mankind (and womankind)\r\n-Do you know how many children die in Africa? Well, going to this thing can change everything.\r\n-(Sob on your knees) Please! Please! I won't live to see another day if you don't take me there.\r\n-Dad, can I mow the lawn? In return you will take me to the thing I told you about.\r\n-Did you know that parents who travel 400 miles in the middle of January are generally more responsible?\r\n\r\nAnyone of those should work :roll: .", + "Solution_11": "since i have yet to gain my super sonic flying abilities, i'll have to buy it when it is on dvd\r\n\r\nlooks really good :D", + "Solution_12": "That is beyond cool!", + "Solution_13": "I think the Chinese did a brain test to see the difference between an average person and an IMOer.\r\nI'm not sure because I'm not Chinese but I heard about this.\r\n\r\nAnyways cheers to Tedrick and Arnav (although Tedrick wasn't in the film..)", + "Solution_14": "[quote=\"8parks11\"]I think the Chinese did a brain test to see the difference between an average person and an IMOer.\nI'm not sure because I'm not Chinese but I heard about this.\n\nAnyways cheers to Tedrick and Arnav (although Tedrick wasn't in the film..)[/quote]\r\n\r\nActually he did appear in the clip, but didn't say anything.", + "Solution_15": "Good choice of music\r\n\r\n<------------------", + "Solution_16": "[quote=\"carpo\"]Good choice of music\n\n<------------------[/quote]\r\nnah should have been \"peace sells\" or something\r\nthat would have been really cool", + "Solution_17": "[quote=\"junggi\"][quote=\"carpo\"]Good choice of music\n\n<------------------[/quote]\nnah should have been \"peace sells\" or something\nthat would have been really cool[/quote]\r\n\r\nbut who's buying?" +} +{ + "Tag": [ + "ratio" + ], + "Problem": "On a mercury thermometer with equally space temperatures from $ \\minus{}30^{\\circ}$ to $ 120^{\\circ}$, the distance between the marks for $ \\minus{}30^{\\circ}$ and $ 120^{\\circ}$ is 20 cm. When the temperature rises from $ \\minus{}8^{\\circ}$ to $ 22^{\\circ}$, what is the distance that the mercury rises?\n\n[asy]draw((0,0)--(15,0)--(15,25)--(0,25)--(0,0)--cycle,linewidth(2));\nfilldraw(circle((3,5),2),black);\nfilldraw((3-sqrt(3),4)--(3+sqrt(3),4)--(3+sqrt(3),20)--(3-sqrt(3),20)--(3-sqrt(3),4)--cycle,black);\ndraw((3-sqrt(3),20)--(3+sqrt(3),20)--(3+sqrt(3),23)--(3-sqrt(3),23)--(3-sqrt(3),20)--cycle,linewidth(1));\n\nfor(int i = 4; i < 21; ++i)\n{\n\tdraw((5,i)--(6,i));\n}\n\nlabel(\"$-30^{\\circ}$\",(6,4),E);\nlabel(\"$120^{\\circ}$\",(6,20),E);\n\ndraw((16.5,4)--(17.5,4),linewidth(1));\ndraw((16.5,20)--(17.5,20),linewidth(1));\ndraw((17,4)--(17,20),linewidth(1));\n\nlabel(\"20 cm\",(17,12),E);[/asy]", + "Solution_1": "Let $ d\\equal{}\\text{distance}.$ We have the ratio $ \\frac{20\\text{ cm}}{150^{\\circ}}\\equal{}\\frac{d}{30^{\\circ}}\\implies d\\equal{}\\frac{20\\text{ cm}}{5}\\equal{}\\boxed{4\\text{ cm}}.$" +} +{ + "Tag": [ + "projective geometry", + "geometry", + "perpendicular bisector", + "Menelaus" + ], + "Problem": "Disjoint circles $ o_1, o_2$, with centers $ I_1, I_2$ respectively, are tangent to the line $ k$ at $ A_1, A_2$ respectively and they lie on the same side of this line. Point $ C$ lies on segment $ I_1I_2$ and $ \\angle A_1CA_2 \\equal{} 90^{\\circ}$. Let $ B_1$ be the second intersection of $ A_1C$ with $ o_1$, and let $ B_2$ be the second intersection of $ A_2C$ with $ o_2$. Prove that $ B_1B_2$ is tangent to the circles $ o_1, o_2$.", + "Solution_1": "We use the degenerate Pappus theorem. Since $ I_1,C,I_2$ are collinear, then the parallel through $ I_1$ to $ CA_2,$ (perpendicular to CA_1) and the parallel through $I_2$ to $CA_1$ (perpendicular to CA_2) concur at a point $ Q$ lying on $ A_1A_2.$ Let $ \\tau$ denote the tangent to $ (I_1)$ through $ B_1.$ Note that $ \\tau$ passes through $ Q,$ because $ CQ$ is the perpendicular bisector of $\\overline{ A_1B_1}$ $\\Longrightarrow$ $ QI_2$ is the Q-external bisector of the isosceles triangle $\\triangle QB_1A_1$ $\\Longrightarrow$ $( I_2)$ is tangent to $\\tau.$", + "Solution_2": "For an alternative approach, see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=223789]Common polar with tangents[/url]\r\n\r\nKostas Vittas.", + "Solution_3": "Can anybody explain me how to conclude that Q lies on A1A2 from Pappus Theorem?", + "Solution_4": "@above since $A_1I_1\\parallel A_2I_2$ and $QI_1\\parallel CA_2,QI_2\\parallel CA_1$ so $A_1I_1\\cap A_2I_2,QI_1\\cap CA_2,QI_2\\cap CA_1$ lie at the line at infinity so By converse of pappus theorem we have $A_2-Q-A_1$ collinear", + "Solution_5": "Let $D_i$ be an intersection of line $A_iI_i$ with $o_i$ for $i \\in {1, 2}$. Let $E$ be an intersection of lines $I_1I_2$ and $B_2D_2$.\nLet $X_1, X_2$ lie on line $B_1B_2$ in such a way that $X_1B_1D_1$ and $X_2B_2A_2$ are acute angles.\n \n$\\angle B_2D_2A_2 = \\angle CA_2A_1 =180^{\\circ} -\\angle A_1CA_2 - \\angle CA_1A_2 = 180^{\\circ} - 90^{\\circ} - \\angle CA_1A_2 = 90^{\\circ} - \\angle CA_1A_2 = \\angle B_1A_1D_1$\n$\\angle B_1A_1D_1 = \\angle B_2D_2A_2$ and $\\angle A_1B_1D_1 = \\angle A_2B_2D_2 = 90^{\\circ}$, so $\\Delta A_1B_1D_1 \\sim \\Delta A_2B_2D_2$.\n$ \\angle CA_2A_1 = \\angle B_1A_1D_1$ and $\\angle A_1CA_2 = A_1B_1D_1 = 90^{\\circ}$, so $\\Delta A_1A_2C \\sim \\Delta A_1B_1D_1$, so\n$\\Delta A_1B_1D_1 \\sim \\Delta A_2B_2D_2 \\sim \\Delta A_1A_2C$\n\n$A_1B_1 \\perp A_2C$ and $B_2D_2 \\perp A_2C$ so $A_1B_1||B_2D_2$.\n$A_1B_1 || B_2D_2$, so $\\angle A_1CI_1 = \\angle D_2EI_1$.\n$\\angle A_1CI_1 = \\angle D_2EI_1$ and $\\angle B_1A_1D_1 = \\angle B_2D_2A_2$ so $\\Delta A_1CI_1 \\sim \\Delta D_2EI_1$.\n\n$\\Delta A_1B_1D_1 \\sim \\Delta A_2B_2D_2$, so $\\frac{|A_1D_1|}{|A_2D_2|} = \\frac{|A_1B_1|}{|B_2D_2|}$\n$\\Delta A_1CI_1 \\sim \\Delta D_2EI_1$, so $\\frac{|A_1I_1|}{|I_2D_2|} = \\frac{|A_1C|}{|ED_2|}$.\n$|A_1I_1| = \\frac{1}{2}|A_1D_1|$ and $|I_2D_2| = \\frac{1}{2}|A_2D_2|$, so $\\frac{|A_1B_1|}{|B_2D_2|} = \\frac{|A_1D_1|}{|A_2D_2|} = \\frac{|A_1I_1|}{|I_2D_2|} = \\frac{|A_1C|}{|ED_2|}$.\n$ \\frac{|A_1B_1|}{|B_2D|} = \\frac{|A_1C|}{|ED_2|} $, so $\\frac{|A_1C|}{|B_1C|} = \\frac{|D_2E|}{|B_2E|} $.\n\nFrom Menelaus theorem for $\\Delta A_2B_2D_2$ and line $CI_2$: $\\frac{|A_2C|}{|CB_2|}\\cdot\\frac{|B_2E|}{|ED_2|}\\cdot\\frac{|D_2I_2|}{|I_2A_2|} =1$.\n$|D_2I_2| = |I_2A_2|$, so $\\frac{|A_2C|}{|CB_2|}\\cdot\\frac{|B_2E|}{|ED_2|} = 1$\n$\\frac{|A_2C|}{|CB_2|}\\cdot\\frac{|B_2E|}{|ED_2|} = 1$ and $\\frac{|A_1C|}{|B_1C|} = \\frac{|D_2E|}{|B_2E|} $ so $\\frac{|A_2C|}{|CB_2|}\\cdot\\frac{|B_1C|}{|A_1C|} = 1$, so $\\frac{|A_2C|}{|CB_2|} = \\frac{|A_1C|}{|B_1C|} $, so $\\Delta B_1CB_2 \\sim \\Delta A_1CA_2$, so\n$\\Delta B_1CB_2 \\sim \\Delta A_1CA_2 \\sim \\Delta A_2B_2D_2 \\sim \\Delta A_1B_1D_1$.\n\n$\\Delta B_1CB_2 \\sim \\Delta A_2B_2D_2$, so $\\angle CB_2B_1 = \\angle B_2D_2A_2$, so $ \\angle X_2B_2A_2 = \\angle B_2D_2A_2$, so $B_1B_2$ is tangent to $o_2$.\n$B_2C \\perp A_1B_1$ and $B_1D_1 \\perp A_1B_1$, so $B_2C || B_1D_1$, so $ \\angle B_1B_2C = X_1B_1D_1$. \n$\\angle B_1B_2C = \\angle B_2D_2A_2 = \\angle B_1A_1D_1$ and $ \\angle B_1B_2C = \\angle X_1B_1D_1$ so $\\angle B_1A_1D_1 = \\angle X_1B_1D_1$, so $B_1B_2$ is tangent to $o_1$." +} +{ + "Tag": [ + "vector", + "trigonometry", + "analytic geometry", + "trig identities", + "Law of Cosines" + ], + "Problem": "Suppose the three vectors $\\overrightarrow a,\\ \\overrightarrow b,\\ \\overrightarrow c$ on a plane satisfy the condition that $|\\overrightarrow a|=|\\overrightarrow b|=|\\overrightarrow c|=|\\overrightarrow a+\\overrightarrow b|=1,\\ \\overrightarrow c$ is perpendicular to $\\overrightarrow a$ and $\\overrightarrow b\\cdot \\overrightarrow c>0.$\r\n\r\n(1) Find the angle forming by$2\\overrightarrow a+\\overrightarrow b$ and $\\overrightarrow b.$\r\n(2) Express the vector $\\overrightarrow c$ in terms of $\\overrightarrow a$ and $\\overrightarrow b.$\r\n\r\n(3)For real numbers $x,y,$ the vector $\\overrightarrow p=x\\overrightarrow a+y\\overrightarrow c$ satisfies the condition that $0\\leq \\overrightarrow p\\cdot \\overrightarrow a\\leq 1,\\ 0\\leq \\overrightarrow p\\cdot \\overrightarrow b\\leq 1.$\r\nFind the maximum value of $\\overrightarrow p\\cdot \\overrightarrow c,$ then express the $\\overrightarrow p$ which gives the maximum value in terms of $\\overrightarrow a$ and $\\overrightarrow b.$", + "Solution_1": "You can just use \\vec. It's shorter to type. $\\vec{a}$ (you can do it your way as well, just wanted to let you know of the \\vec in case you didn't)", + "Solution_2": "Thank you, joml88.\r\n\r\nkunny", + "Solution_3": "No solution but just curious, what colleges still give entrance exams?", + "Solution_4": "The \"Center exam\" is actually a standardized test that many Japanese 12th graders need to take in order to go to college. It is sort of similar to SAT/ACT in the U.S., but it's harder. (Besides, the test is offered only once a year.)\r\n\r\nMany colleges also have their own entrance exams, so students often have to take both the Center exam and the colleges' own exams.", + "Solution_5": "Is the answer of problem one $\\arccos\\frac{\\sqrt{3}}{3}$.", + "Solution_6": "[quote=\"shobber\"]Is the answer of problem one $\\arccos\\frac{\\sqrt{3}}{3}$.[/quote]\r\n\r\nby constructing the diagram , I found the angle to be $90^{\\circ}$", + "Solution_7": "[hide=\"2\"]\nWe know that $|\\vec{a}+\\vec{b}|=1$ like a and b, and thus, by the Law of Cosines (if A and B were to form a triangle), the angle between $\\vec{a}$ and $\\vec{b}$ is $120^{\\circ}$. So from there, we find that \\[ \\vec{c}=\\frac{2}{\\sqrt{3}}\\vec{b}+\\frac{1}{\\sqrt{3}}\\vec{a} \\][/hide]\n\nNo time to do 3, but i'll say that [hide] $\\vec{p}\\cdot \\vec{a}=x$[/hide]\n\nFor 1, [hide=\"1\"] set up a coordinate system, and get $\\vec{b}=(-\\frac{1}{2}, \\frac{\\sqrt{3}}{2})$, $\\vec{a}=(1, 0)$. Thus, $2\\vec{a}+\\vec{b}=(\\frac{3}{2}, \\frac{\\sqrt{3}}{2})$ Then use the dot product to get \\[ (2\\vec{a}+\\vec{b}) \\cdot \\vec{b} =0 \\implies \\cos{\\theta}=0 \\implies \\theta=\\boxed{90^{\\circ}} \\][/hide]", + "Solution_8": "K8107, you actually get four different sets of vectors of $a,b$, but they all give the same dot product.", + "Solution_9": "Sorry for that I made a typo. I have just edited.\r\n\r\nkunny", + "Solution_10": "[quote=\"Kalle\"]K8107, you actually get four different sets of vectors of $a,b$, but they all give the same dot product.[/quote]\r\n\r\nOh yeah, true...$\\vec{b}$ could be $(-\\frac{1}{2}, -\\frac{\\sqrt{3}}{2})$.", + "Solution_11": "[quote=\"shyong\"][quote=\"shobber\"]Is the answer of problem one $\\arccos\\frac{\\sqrt{3}}{3}$.[/quote]\n\nby constructing the diagram , I found the angle to be $90^{\\circ}$[/quote]\r\n\r\nThat's right, shyong, K8107 :)" +} +{ + "Tag": [ + "geometry", + "rectangle" + ], + "Problem": "The magical land of Oz, in the shape of a perfect 200milesX350miles rectangle, is divided into an array of congruent 2milesX2miles squarish kingdoms. A train starts at the SW corner of Oz, and moves on a straight track (of no thickness) towards the NE corner. At every third kingdom it passes through, it unloads a car of coal. At every forth kingdom, it unloads a car of anchovies. At every fifth kingdom it unloads a car of lumber. (a) At how many kingdoms does the train unload a car of each of coal, anchovies, and lumber?\r\n\r\nEDIT: hmm...I meant to say in how many kingdoms does the train unload atleast one car...(b)", + "Solution_1": "[hide]First we have to find the number of kingdoms in which it passes through.\n\nI get 175+100-GCD(175,100)=250.\n\nNow, we find the LCM(3,4,5)=60.\n\nSo we just do the greatest integer less than 250/60 which is 4.\n\n$4$\nAm I right?\n\n[/hide]", + "Solution_2": "[hide]$\\frac{\\frac{70000}{2^2}}{403}-3=40.42$, so about 41 kingdoms.[/hide]\r\n I did this obscure solution in under a minute so I expect it to be wrong :blush: .", + "Solution_3": "[hide]It goes through 250 cities\nSo the answer is $\\frac{250}{60}$ rounded down which equals $\\boxed{4}$[/hide]", + "Solution_4": "[hide]$250/3=83$\n$250/4=62$\n$250/5=50$\n\n$250/12=20$\n$250/20=12$\n$250/15=16$\n\n$250/60=4$\n\nThe answer for (b) is therefore is $83+62+50-20-12-16+4=\\boxed{151}$[/hide]", + "Solution_5": "Major :oops_sign: and :wallbash_red: \r\n[hide]$\\frac{200}{2}+\\frac{350}{2}-25=250$ cities. $250\\div{60}=4.16=\\boxed{4}$ [/hide]", + "Solution_6": ":spam: Any reason you posted the same thing twice? :spam:", + "Solution_7": "[quote=\"SplashD\"]:spam: Any reason you posted the same thing twice? :spam:[/quote] In my defense, I was deleting the post, while you posted the message. Sorry though.", + "Solution_8": "For (b), for each set of 60, there are 20 multiples of 3, 15 multiples of 4, 12 multiples of 5, 5 of 12, 4 of 15, 3 of 20, and 1 of 60. By the Inclusion-Exclusion thing (I can't remember the exact name), it unloads at $4\\times (20+15+12-5-4-3+1)=4\\times (36)=144$ times in the first 240. Then since there are 250 cities (using everyone else's number, too lazy to do it myself) it also unloads at 243, 244, 245, 246, 248, 249, and 250, for a total of $144+7=151$ cities.", + "Solution_9": "[quote=\"SplashD\"][hide]$250/3=83$\n$250/4=62$\n$250/5=50$\n\n$250/12=20$\n$250/20=12$\n$250/15=16$\n\n$250/60=4$\n\nThe answer for (b) is therefore is $83+62+50-20-12-16+4=\\boxed{151}$[/hide][/quote]\r\n\r\nNow you tell me\r\n\r\nThe answer is the way SplashD, did it so I got it.", + "Solution_10": "[hide=\"a\"]\nThe train passes through 250 kingdoms. Every 60th kingdom, the train drops off 3 cars.\n$\\frac{250}{60}=4.1\\overline{6}$\nRounded, the answer is $4$[/hide]" +} +{ + "Tag": [ + "algebra", + "polynomial", + "induction", + "calculus", + "integration", + "logarithms", + "function" + ], + "Problem": "Let $ a$, $ b$, $ c$ be three real numbers. For each positive integer number $ n$, $ a^n \\plus{} b^n \\plus{} c^n$ is an integer number. Prove that there exist three integers $ p$, $ q$, $ r$ such that $ a$, $ b$, $ c$ are the roots of the equation $ x^3 \\plus{} px^2 \\plus{} qx \\plus{} r \\equal{} 0$.", + "Solution_1": "[I think you forgot an $ r$ in the last equation]\r\nThis a special case of a pretty well known subject (linear recurrene sequences)... \r\nAnyway, put $ x_n \\equal{} a^n \\plus{} b^n \\plus{} c^n$. $ x_n$ satisfies a linear recurrence: $ (*)$ $ x_{n \\plus{} 3} \\equal{} \\lambda_1x_{n \\plus{} 2} \\plus{} \\lambda_2x_{n \\plus{} 1} \\plus{} \\lambda_3x_{n}$, where $ \\lambda_{1,2,3}$ satisfy: $ x^3 \\minus{} \\lambda_1x^2 \\minus{} \\lambda_2x \\minus{} \\lambda_3 \\equal{} (x \\minus{} a)(x \\minus{} b)(x \\minus{} c)$ $ (**)$.\r\nWe want to show that $ \\lambda_{1,2,3}$ are integers. But it follows from the fact that $ x_{n}$ are integers: we can get the value of $ \\lambda_{i}$ from the system $ x_{n \\plus{} 3} \\equal{} \\lambda_1x_{n \\plus{} 2} \\plus{} \\lambda_2x_{n \\plus{} 1} \\plus{} \\lambda_3x_{n}, n \\equal{} 1,2,3$. It implies that $ \\lambda_{i}$ are rational.\r\nAnd now I am a bit stuck... We should show that a linear recurrence sequence with rational coefficients (at least one being non-integer), whose characteristic polynomial has distinct roots, has some non-integer terms.\r\n\r\nEDIT: It is equivalent to $ \\sum a, \\sum ab, abc$ being integers (by comparing coefficients in $ (**)$). \r\n$ a \\plus{} b \\plus{} c$ is integer by plugging $ n \\equal{} 1$ in the given property. \r\n$ 2(ab \\plus{} bc \\plus{} ca) \\equal{} (a \\plus{} b \\plus{} c)^2 \\minus{} \\sum a^2$, so $ \\sum ab$ is either an integer or half an integer. \r\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca)$, so $ abc$ is also either an integer or half an integer.\r\nFirst case: $ \\lambda_{2}$ is the only non-integer. Thus, $ (*)$ implies that $ v_2(x_{n \\plus{} 3}) \\equal{} v_2(x_{n \\plus{} 1}) \\minus{} 1$, which is a contradiction for sufficiently large $ n$.\r\nSecond and last case: $ \\lambda_{2,3}$ are both half-integers. Thus, $ (*)$ implies that $ v_2(x_{n \\plus{} 3}) \\equal{} \\text{min} (v_2(x_{n \\plus{} 1}),v_2(x_{n})) \\minus{} 1$, which is a contradiction for sufficiently large $ n$.\r\n\r\nWe can also talk about Newton's sums instead of the general theory of recurrence sequences. It talks exactly about the expressions $ a^n\\plus{}b^n\\plus{}c^n$.", + "Solution_2": "[quote=\"bambaman\"]\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca)$, so $ abc$ is also either an integer or half an integer.[/quote]\r\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{}3abc \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca)$", + "Solution_3": "You're right... I am not sure how to correct my proof. Does someone want to do that? Anyway, I think there's a more intuitive proof, I will try to find it.", + "Solution_4": "I have a schema in my head to reach final solution, let's use the following wellknown lemma ( i rembere that could be proved by induction)\r\n\r\nLemma 1:\r\n\r\nIf we have c1,c2,c3,..., ck are different rational number greater than 1. Let a1,.., ak be constant. Suppose that there exist a sequence of integer Q_n so that\r\n\r\na1*c1^n+a2*c2^n+.........+ak*ck^n - Q_n tends to zero. Then c1,..,ck must be integer number\r\n\r\nLemma 2:\r\n\r\nThe sequence {Xn} with by X0=a, X1=b, both a and b are integer.\r\n\r\nX_(n+1)=D.X_n - X_(n-1)\r\n\r\nElement of this sequence are all integers if and only if:\r\n\r\n1. If a does not divides b, then D must be an integer\r\n2. If a divides b, then D is either an integer or D=(a/b+b/a)", + "Solution_5": "We can show $ A \\equal{} 6abc$ and $ B \\equal{} 6a^2b^2c^2$ are all integers.\r\nHence $ B \\equal{} \\frac {A^2}6$ implies $ 6|A$ and $ abc$ is integer.\r\n\r\nWe can show $ C \\equal{} 2(ab \\plus{} ac \\plus{} bc)$ and $ D \\equal{} 2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)$ are integers.\r\nThen $ \\frac {C^2}2 \\equal{} D \\plus{} 4abc(a \\plus{} b \\plus{} c)$ is integer thus $ 2|C$ and $ ab \\plus{} ac \\plus{} bc \\equal{} \\frac C2$ is integer.", + "Solution_6": "Let $ P(x) \\equal{} (x \\minus{} a)(x \\minus{} b)(x \\minus{} c) \\equal{} x^3 \\plus{} px^2 \\plus{} qx \\plus{} r$. We are given that\r\n\r\n$ \\frac{P'(x)}{P(x)} \\equal{} \\sum_{n \\ge 0} (a^n \\plus{} b^n \\plus{} c^n) x^n$\r\n\r\nhas all integer coefficients. It follows that the integral\r\n\r\n$ \\ln P(x) \\plus{} C \\equal{} \\sum_{n \\ge 0} \\frac{a^n \\plus{} b^n \\plus{} c^n}{n\\plus{}1} x^{n\\plus{}1}$\r\n\r\nhas integer coefficients when viewed as an exponential generating function. $ x \\equal{} 0$ gives $ \\ln r \\plus{} C \\equal{} 0 \\Leftrightarrow C \\equal{} \\minus{} \\ln r$, hence the exponential\r\n\r\n$ \\frac{x^3 \\plus{} px^2 \\plus{} qx \\plus{} r}{r}$\r\n\r\nhas integer coefficients, again when viewed as an exponential generating function. (This is an easy consequence of the fundamental theorem of exponential generating functions, and can also be proven directly.) In particular, $ \\frac{q}{r}$ is an integer, $ \\frac{p}{r}$ is half an integer, and $ \\frac{1}{r}$ is a sixth of an integer. I think an argument along xxp2000's lines finishes it.", + "Solution_7": "[quote=\"xxp2000\"]We can show $ A \\equal{} 6abc$ and $ B \\equal{} 6a^2b^2c^2$ are all integers.\nHence $ B \\equal{} \\frac {A^2}6$ implies $ 6|A$ and $ abc$ is integer.\n\nWe can show $ C \\equal{} 2(ab \\plus{} ac \\plus{} bc)$ and $ D \\equal{} 2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)$ are integers.\nThen $ \\frac {C^2}2 \\equal{} D \\plus{} 4abc(a \\plus{} b \\plus{} c)$ is integer thus $ 2|C$ and $ ab \\plus{} ac \\plus{} bc \\equal{} \\frac C2$ is integer.[/quote]\r\nThis solution is similar to my solution in the contest :D", + "Solution_8": "[quote=\"April\"]Let $ a$, $ b$, $ c$ be three real numbers. For each positive integer number $ n$, $ a^n \\plus{} b^n \\plus{} c^n$ is an integer number. Prove that there exist three integers $ p$, $ q$, $ r$ such that $ a$, $ b$, $ c$ are the roots of the equation $ x^3 \\plus{} px^2 \\plus{} qx \\plus{} r \\equal{} 0$.[/quote]\r\n\r\nMy proof: From $ a\\plus{}b\\plus{}c,a^2\\plus{}b^2\\plus{}c^2,a^4\\plus{}b^4\\plus{}c^4 \\in Z$ we have $ 2(ab\\plus{}bc\\plus{}ca),2(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2) \\in Z$\r\nThen,from the formula: $ x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz\\equal{}(x\\plus{}y\\plus{}z)(x^2\\plus{}y^2\\plus{}z^2\\minus{}xy\\minus{}yz\\minus{}zx)$, apply for $ (x,y,z)\\equal{}(a,b,c)$ and $ (a^2,b^2,c^2)$ we have:\r\n$ 6abc \\in Z$ and $ 6a^2b^2c^2 \\in Z$\r\nSet $ 6abc\\equal{}k \\rightarrow 6a^2b^2c^2\\equal{}\\frac{k^2}{6} \\rightarrow 6|k$ and $ abc \\in Z$\r\nFrom $ 2(ab\\plus{}bc\\plus{}ca)^2\\equal{}2(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)\\plus{}4abc(a\\plus{}b\\plus{}c)$ we have $ 2(ab\\plus{}bc\\plus{}ca)^2 \\in Z$\r\nSet $ 2(ab\\plus{}bc\\plus{}ca)\\equal{}q \\rightarrow 2(ab\\plus{}bc\\plus{}ca)^2 \\equal{} \\frac{q^2}{2}$ so $ 2|q$ and $ ab\\plus{}bc\\plus{}ca \\in Z$\r\n\r\nWe are done!", + "Solution_9": "I have edited my post.\n\nSince $a^n+b^n+c^n$ all are integer then if $a$,$b$,$c$ all can be irrational in same time,then if $a,b$ irrational then they are conjugate.\n[b]Case 1:[/b]\nLet, $a=\\frac{p_1}{q_1}$, $b=\\frac{p_2}{q_2}$ ,$c=\\frac{p_3}{q_3}$\n\nthen $(a+b+c)^2-(a^2+b^2+c^2)=\\frac{(p_1q_2q_3+q_1p_2q_3+q_1q_2p_3)^2-(p_1q_2q_3)^2-(q_1p_2q_3)^2-(q_1q_2p_3)^2}{q_1^2q_2^2q_3^2}$. Since; g.c.d.$(p_i,q_i)=1$ then if any $q_1,q_2,q_3$ are divisible by 2 then $p_i$ are not.\nHence $(p_1q_2q_3+q_1p_2q_3+q_1q_2p_3)^2-(p_1q_2q_3)^2-(q_1p_2q_3)^2-(q_1q_2p_3)^2$ is divisible by 4.\n\n$\\frac{(p_1q_2q_3+q_1p_2q_3+q_1q_2p_3)^n-(p_1q_2q_3)^n-(q_1p_2q_3)^n-(q_1q_2p_3)^n}{q_1^nq_2^nq_3^n}=P$\n\nthen $q_i|q_{i+1}q_{i+2}$, or,$q_ik_i=q_{i+1}q_{i+2}$ ,or,$k_1k_2k_3=q_1q_2q_3$, or,$k_i\\nmid{q_{i+1}}$ & $k_i\\nmid{q_{i+2}}$ ,then $k_i|q_i$, then $q_i^2=q{i+1}q{i+2}\\Rightarrow{q_1q_2q_3=1}$ .Hence $abc=p_1p_2p_3$\n\nSince $a+b+c$,$ab+bc+ca$,$abc$ are integers.$a+b+c=-p$,$ab+bc+ca=q$,$abc=-r$ satisfies $x^3+px^2+xq+r=0$ ,then $p,q,r$ exist.\n\n[b]Case 2:[/b]\nIf $a,b$ are conjugate, then it should be :$a=s+\\sqrt{r}$ & $b=s-\\sqrt{r}$ where $s$ & $r$ should be integer and $c$ also. then $a+b+c$, $ab+bc+ca$, $abc$ all are integer, this follows $p,q,r$ exist.", + "Solution_10": "[quote=\"MANMAID\"]...then the the irr. part of $a,b$ has n-th root ...[/quote]\nWhat is the \"irrational part\" of a real ?", + "Solution_11": "[quote=\"pco\"][quote=\"MANMAID\"]...then the the irr. part of $a,b$ has n-th root ...[/quote]\nWhat is the \"irrational part\" of a real ?[/quote]\n\nwe can assume a real number $r=\\frac{p}{q}+\\alpha$ , where $\\frac{p}{q}$ is rational & $\\alpha$ is irrational.\nFor example: $10.010010001000010000010000001.......$ ,here we can assume $10$ is rational & $.010010001000010000010000001......$ is irrational.\n$\\frac{p}{q}$ or $\\alpha$ may be zero.\n\nIs it not true.", + "Solution_12": "For any real, we can write $r=a+b$ with $a\\in\\mathbb Q$ and $b\\notin\\mathbb Q$ in infinitely many manners (just transform $(a,b)\\to(a+a',b-a')$ where $a'$ is any rational number).\n\nSo, if you want to establish properties on the so called \"irrational part\" of a real $r$, you must at least define with precision what decomposition in the infinitely many you use. Else, a property true for one decomposition will not be true for another one.", + "Solution_13": "[quote=\"pco\"]For any real, we can write $r=a+b$ with $a\\in\\mathbb Q$ and $b\\notin\\mathbb Q$ in infinitely many manners (just transform $(a,b)\\to(a+a',b-a')$ where $a'$ is any rational number).\n\nSo, if you want to establish properties on the so called \"irrational part\" of a real $r$, you must at least define with precision what decomposition in the infinitely many you use. Else, a property true for one decomposition will not be true for another one.[/quote]\nis the solution that I did is right, \" decomposition in the infinitely many\" what is the meaning of this?\nI just wanted to show $a,b,c$ are not irr.., so I used this property. Thanks .", + "Solution_14": "No you are not right. You can see that you are wrong in assuming $a, b, c$ are rational with the simple example $a=1+\\sqrt{2}, b=1-\\sqrt{2}, c=0$. Then $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=14$, and $a^n+b^n+c^n=2(a^{n-1}+b^{n-1}+c^{n-1})+(a^{n-2}+b^{n-2}+c^{n-2})$ is also an integer. So it's an integer for all $n$. But $1+\\sqrt{2}$ is not rational.", + "Solution_15": "[quote=\"tenniskidperson3\"]No you are not right. You can see that you are wrong in assuming $a, b, c$ are rational with the simple example $a=1+\\sqrt{2}, b=1-\\sqrt{2}, c=0$. Then $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=14$, and $a^n+b^n+c^n=2(a^{n-1}+b^{n-1}+c^{n-1})+(a^{n-2}+b^{n-2}+c^{n-2})$ is also an integer. So it's an integer for all $n$. But $1+\\sqrt{2}$ is not rational.[/quote]\nthough it is right (sorry for my wrong argument) you can not prove it for $a+b^{\\frac{1}{n}}$ for $n\\neq{2}$ or any other irrational." +} +{ + "Tag": [], + "Problem": "Hi everyone. I am newcomer in Canada :lol:. I need TOEF for university :) . I live in Scarbourgh. Where I can do test TOEF ?", + "Solution_1": "do you mean toefl? (test of english as a foreign language?)", + "Solution_2": "Google and I also think he means toefl : http://www.google.ca/search?hl=en&q=TOEF&btnG=Google+Search&meta=", + "Solution_3": "Sure. I need TOEFL for university :). I think I do test TOEFL in Downtown. But I don know exactly where :(.", + "Solution_4": "just go to this site: http://www.ets.org/toefl/\r\nregister for the test and choose a location" +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "Given $ a,b,c >0$, and $ a\\plus{}b\\plus{}c\\equal{}3$.Prove that:\r\n$ \\frac{a}{9\\plus{}a\\minus{}c}\\plus{}\\frac{b}{9\\plus{}b\\minus{}a}\\plus{}\\frac{c}{9\\plus{}c\\minus{}b} \\le \\frac{11}{6}\\minus{}\\frac{a^2b}{a\\plus{}b}\\minus{}\\frac{b^2c}{b\\plus{}c}\\minus{}\\frac{c^2a}{c\\plus{}a}$\r\n :)", + "Solution_1": "Sorry,but it's not nice any more with me (it looks so long) :wink: \r\nBesides, it's very weak. :maybe: \r\n$ \\frac {a}{9 \\plus{} a \\minus{} c} \\plus{} \\frac {b}{9 \\plus{} b \\minus{} a} \\plus{} \\frac {c}{9 \\plus{} c \\minus{} b} \\le \\frac {1}{3}$ (obviously trues by Cauchy Schwarts)\r\nand $ \\sum\\ \\frac {a^2b}{a \\plus{} b} \\leq\\ \\sum\\ \\frac {\\sqrt {a^3b}}{2} \\leq\\ \\frac {3}{2}$ (by Vasc's ineq)", + "Solution_2": "[quote=\"nguoivn\"] \n$ \\frac {a}{9 \\plus{} a \\minus{} c} \\plus{} \\frac {b}{9 \\plus{} b \\minus{} a} \\plus{} \\frac {c}{9 \\plus{} c \\minus{} b} \\le \\frac {1}{3}$ (obviously trues by Cauchy Schwarts)[/quote]\r\nPlease post solution. \r\nMy solution use the inequality of Hung and Vasc: :wink: \r\n$ \\sum\\ \\frac{a}{4\\minus{}b} \\le 1$", + "Solution_3": "[quote=\"quykhtn-qa1\"][quote=\"nguoivn\"] \n$ \\frac {a}{9 \\plus{} a \\minus{} c} \\plus{} \\frac {b}{9 \\plus{} b \\minus{} a} \\plus{} \\frac {c}{9 \\plus{} c \\minus{} b} \\le \\frac {1}{3}$ (obviously trues by Cauchy Schwarts)[/quote]\nPlease post solution. \nMy solution use the inequality of Hung and Vasc: :wink: \n$ \\sum\\ \\frac {a}{4 \\minus{} b} \\le 1$[/quote]\r\nI know your solution (also my second proof) but we don't have to use that lenma, it's very easy :wink: \r\nWe can write it into: $ \\frac {a}{4a \\plus{} 3b \\plus{} 2c} \\leq\\ \\frac {1}{3}$\r\n<=> $ \\sum\\ \\frac {3b \\plus{} 2c}{4a \\plus{} 3b \\plus{} 2c} \\geq\\ \\frac {5}{3}$\r\nBy Cauchy Schwarts, $ LHS \\geq\\ \\sum\\ \\frac {25(a \\plus{} b \\plus{} c)^2}{\\sum\\ (3b \\plus{} 2c)(4a \\plus{} 3b \\plus{} 2c)} \\geq\\ \\frac {5}{3}$", + "Solution_4": "[quote=\"nguoivn\"][quote=\"quykhtn-qa1\"][quote=\"nguoivn\"] \n$ \\frac {a}{9 \\plus{} a \\minus{} c} \\plus{} \\frac {b}{9 \\plus{} b \\minus{} a} \\plus{} \\frac {c}{9 \\plus{} c \\minus{} b} \\le \\frac {1}{3}$ (obviously trues by Cauchy Schwarts)[/quote]\nPlease post solution. \nMy solution use the inequality of Hung and Vasc: :wink: \n$ \\sum\\ \\frac {a}{4 \\minus{} b} \\le 1$[/quote]\nI know your solution (also my second proof) but we don't have to use that lenma, it's very easy :wink: \nWe can write it into: $ \\frac {a}{4a \\plus{} 3b \\plus{} 2c} \\leq\\ \\frac {1}{3}$\n<=> $ \\sum\\ \\frac {3b \\plus{} 2c}{4a \\plus{} 3b \\plus{} 2c} \\geq\\ \\frac {5}{3}$\nBy Cauchy Schwarts, $ LHS \\geq\\ \\sum\\ \\frac {25(a \\plus{} b \\plus{} c)^2}{\\sum\\ (3b \\plus{} 2c)(4a \\plus{} 3b \\plus{} 2c)} \\geq\\ \\frac {5}{3}$[/quote]\r\nNice solution :)" +} +{ + "Tag": [ + "geometry", + "parallelogram", + "trigonometry", + "circumcircle", + "geometry solved" + ], + "Problem": "Let $ABCD$ be a parallelogram. Suppose a line passing through $C$ and lying outside the parallelogram meets $AB$ and $AD$ produced at $E$ and $F$ respectively. Show that \\[ AC^2 + CE \\cdot CF = AB \\cdot AE + AD \\cdot AF . \\]", + "Solution_1": "Denote $\\alpha = \\angle DAB,\\ \\beta = \\angle ABC$ the parallelogram internal angles, $\\alpha + \\beta = 180^\\circ$. Using the cosine theorem, for example, for the triangle $\\triangle FDC$ together with $AB = DC$, we get\r\n\r\n$CF = DC^2 + DF^2 - 2\\ DC \\cdot DF \\cos \\alpha = AB^2 + DF^2 - 2\\ AB \\cdot DF\\ \\cos \\alpha$\r\n\r\n$1= \\left(\\frac{AB}{CF}\\right)^2 + \\left(\\frac{DF}{CF}\\right)^2 - 2\\ \\frac{AB}{CF} \\cdot \\frac{DF}{CF}\\ \\cos \\alpha$\r\n\r\nThe triangles $\\triangle FDC \\sim \\triangle CBE$, because their corresponding side lines are parallel: $DC \\parallel AB \\equiv BE,\\ DF \\equiv AD \\parallel BC$ and $CF \\equiv CF$. Hence,\r\n\r\n$\\frac{AB}{CF} = \\frac{DC}{CF} = \\frac{BE}{CE}$\r\n\r\n$\\frac{DF}{CF} = \\frac{BC}{CE} = \\frac{AD}{CE}$\r\n\r\nSubstituting appropriately (i.e., not everywhere) for $\\frac{AB}{CF},\\ \\frac{DF}{CF}$ into the last cosine theorem formula,\r\n\r\n$1 = \\frac{AB}{CF} \\cdot \\frac{BE}{CE} + \\frac{AD}{CE} \\cdot \\frac{DF}{CF} - 2\\ \\frac{AB}{CF} \\cdot \\frac{AD}{CE}\\ \\cos \\alpha$\r\n\r\nMultiplying by the product $CE \\cdot CF$ and substituting for $BE = AE - AB,\\ DF = AF - AD$,\r\n\r\n$CE \\cdot CF = AB \\cdot (AE - AB) + AD \\cdot (AF - AD) - 2\\ AB \\cdot AD\\ \\cos \\alpha$\r\n \r\n$CE \\cdot CF = AB \\cdot AE + AD \\cdot AF - (AB^2 + AD^2 + 2\\ AB \\cdot AD\\ \\cos \\alpha)$\r\n \r\nFinally, using the cosine theorem for the triangle $\\triangle ABC$ together with $AD = BC,\\ \\cos \\alpha = -\\cos \\beta$, we get\r\n\r\n$AC^2 = AB^2 + BC^2 - 2\\ AB \\cdot BC\\ \\cos \\beta = AB^2 + AD^2 + 2\\ AB \\cdot AD\\ \\cos \\alpha$\r\n\r\nSubstituting this to the previous equation yields the desired result:\r\n\r\n$CE \\cdot CF = AB \\cdot AE + AD \\cdot AF - AC^2$", + "Solution_2": "[quote=\"Rushil\"]Let $ABCD$ be a parallelogram. Suppose a line passing through $C$ and lying outside the parallelogram meets $AB$ and $AD$ produced at $E$ and $F$ respectively. Show that \\[ AC^2 + CE \\cdot CF = AB \\cdot AE + AD \\cdot AF . \\][/quote]\r\n\r\nLet's try to find a solution hiding the fact that this problem is nothing but a reformulation of Stewart's theorem...\r\n\r\nLet the line AC meet the circumcircle of triangle AEF at G (apart from A). Then, by the intersecting chords theorem, $CE\\cdot CF=AC\\cdot CG$.\r\nLet the line AC meet the circumcircle of triangle EBC at U (apart from C). Then, by the intersecting chords theorem, $AB\\cdot AE=AU\\cdot AC$.\r\nLet the line AC meet the circumcircle of triangle FCD at V (apart from C). Then, by the intersecting chords theorem, $AD\\cdot AF=AV\\cdot AC$.\r\n\r\nHence, the equation that we have to prove, $AC^2 + CE \\cdot CF = AB \\cdot AE + AD \\cdot AF$, becomes $AC^2+AC\\cdot CG=AU\\cdot AC+AV\\cdot AC$. Division by AC transforms this into AC + CG = AU + AV.\r\n\r\nNow, we will show that AU = VG. This will yield AC + CG = AG = VG + AV = AU + AV, and thus the problem will be solved.\r\n\r\nThe rest of the solution is angle chasing, so we will use directed angles modulo 180\u00b0.\r\n\r\nSince the points E, B, C and U lie on one circle, < CUB = < CEB and < UBE = < UCE. In other words, < AUB = - < AEC and < UBA = - < ECA. Hence, the triangles AUB and AEC are oppositely similar, so that $\\frac{AU}{AE}=\\frac{AB}{AC}$.\r\nSince the quadrilateral ABCD is a parallelogram, BC || AD, so that < EAF = < EBC. Since the points A, E, F and G lie on one circle, < EFG = < EAG and < EGF = < EAF. In other words, < EFG = - < CAB and < EGF = - < CBA (in fact, since < EAF = < EBC, the equation < EGF = < EAF becomes < EGF = < EBC = - < CBA). Hence, the triangles FGE and ABC are oppositely similar, so that $\\frac{AB}{AC}=\\frac{FG}{FE}$.\r\nSince the points F, C, D and V lie on one circle, < CVF = < CDF. Since the quadrilateral ABCD is a parallelogram, CD || AB, and thus < CDF = < EAF. Hence, < GVF = < CVF = < CDF = < EAF. Also, since the points A, E, F and G lie on one circle, < AGF = < AEF, what is equivalent to < VGF = < AEF. From < VGF = < AEF and < GVF = < EAF, it follows that the triangles VGF and AEF are directly similar, so that $\\frac{FG}{FE}=\\frac{VG}{AE}$.\r\n\r\nHence, $\\frac{AU}{AE}=\\frac{AB}{AC}=\\frac{FG}{FE}=\\frac{VG}{AE}$, so that AU = VG, and we are done.\r\n\r\n darij", + "Solution_3": "$BC\\parallel AF\\Longrightarrow \\frac{AE}{AB}=\\frac{FE}{FC}\\Longrightarrow AB=\\frac{AE\\cdot FC}{EF}\\ \\ (1)\\ .$\r\n\r\n$CD\\parallel AE\\Longrightarrow \\frac{AF}{AD}=\\frac{EF}{EC}\\Longrightarrow AD=\\frac{AF\\cdot EC}{EF}\\ \\ (2)\\ .$\r\n\r\nThe [u]Stewart's theorem[/u] in the triangle $AEF$ for the cevian-line $AC\\Longrightarrow$\r\n\r\n$AC^2\\cdot EF+EF\\cdot EC\\cdot FC=AE^2\\cdot CF+AF^2\\cdot CE\\Longrightarrow$\r\n\r\n$AC^2+CE\\cdot CF=AE\\cdot \\frac{AE\\cdot FC}{EF}+AF\\cdot \\frac{AF\\cdot EC}{EF}\\ .$\r\n\r\nFrom the relations $(1)$ and $(2)$ results: $AC^2+CE\\cdot CF=AB\\cdot AE+AD\\cdot AF\\ .$", + "Solution_4": "By Thales theorem on \u2206 EAF, BC|| AD\nEB/ AB=CE/ CF\nAE/ AB=EF/ CF( ADD 1 O BOTH SIDES)....(1)\nBy Thales theorem on \u2206 EAF, DC||AE\nED/ AD=CF/ CE\nAF/ AD=EF/ CE( ADD 1 O BOTH SIDES)....(2)\nBy Stewart theorem in \u2206 EAF,\nCE.CF.EF+AC\u00b2.EF=AF\u00b2.CE+AE\u00b2.CF\nAC\u00b2+CE.CF=AF\u00b2(CE/EF)+ AE\u00b2(CF/EF)\nAC\u00b2+CE.CF=AF\u00b2.(AD/ AF)+ AE\u00b2(AB/AE)[ From (1)&(2)]\nAC\u00b2+CE.CF=AD.AF+ AB.AE( proved)\n# Krishijivi \n\n" +} +{ + "Tag": [ + "function", + "inequalities", + "integration", + "calculus", + "derivative", + "real analysis", + "real analysis unsolved" + ], + "Problem": "If a function $ f$ statisfy $ f'(x) \\leq f'(x\\plus{}\\frac {1}{n})$ for any real value $ x$ and natural number $ n$, prove that $ f$ is continuously differentiable. Thank you for the solution of the problem before.", + "Solution_1": "The inequality itself suggests that f'(x) exists for all real x. Therefore f(x) is differentiable and hence continuous everywhere on the real axis. That makes f continuously differentiable everywhere..\r\n\r\nAm I missing something ? :?: :maybe:", + "Solution_2": "\"Continuously differentiable\" means \"differentiable, with a continuous derivative.\"", + "Solution_3": "Let $ u,v$ be such that $ u \\leq v$ .\r\n$ f'(x)\\leq f'(x\\plus{}\\frac 1n) \\ \\Longrightarrow \\ \\int_u^v f'(x)dx \\leq \\int_u^v f'(x\\plus{}\\frac 1n)dx \\ \\Longrightarrow$$ \\ f(v)\\minus{}f(u)\\leq f(v\\plus{}\\frac 1n)\\minus{}f(u\\plus{}\\frac 1n)\\ \\Longrightarrow\\ \\frac {f(u\\plus{}\\frac 1n)\\minus{}f(u)}{\\frac 1n}\\leq \\frac {f(v\\plus{}\\frac 1n)\\minus{}f(v)}{\\frac 1n}$\r\nWith $ (n\\rightarrow \\infty )\\Longleftrightarrow (\\frac 1n \\rightarrow 0 )$, we obtain $ f'(u)\\leq f'(v)$, i.e. $ f'$ is increasing. But $ f'$ is a derivative function so it has the Darboux property (i.e. Intermediate Value Property) .\r\nFinally $ f'$ is continuous if you know that all monotonic function with the Darboux property is continuous .\r\n :cool:", + "Solution_4": "but when you apply the fundamental theorem of calculus, you need the assumption of the continuity of the function $ f'(x)$. Don't you think so?", + "Solution_5": "That's certanly correct, but the only thing Diogene really needs is that $ x\\mapsto f(x\\plus{}\\tfrac 1n)\\minus{}f(x)$ is increasing and the mean value theorem is enough for that, there is no need to integrate anything :).", + "Solution_6": "[quote=\"thinkagain\"]but when you apply the fundamental theorem of calculus, you need the assumption of the continuity of the function $ f'(x)$. Don't you think so?[/quote]\nNo, I think that the integrability is sufficient ! Just for the fun : Second Fundamental Theorem of Calculus (Riemann integration) \n- Let $ F$ continuous in $ [a,b]$ and $ f$ a Riemann integrable function such that $ F'(x) \\equal{} f(x)$ (except at most in a finite number of points). Then $ F(x) \\minus{} F(a) \\equal{} \\int_a^xf(t)\\,dt$\n(And you can find an other one for Lebesgue integration)\n\n[quote=\"fedja\"]That's certanly correct, but the only thing Diogene really needs is that is increasing $ x \\mapsto f(x \\plus{} \\frac 1n) \\minus{} f(x)$ and the mean value theorem is enough for that, there is no need to integrate anything Smile.[/quote]\r\n\r\nfedja is right (and Diogene is looking for complicated and incorrect proofs :rotfl: ) ,manifestly the integration is not necessary : $ x\\mapsto f(x \\plus{} \\frac 1n) \\minus{} f(x)$ is increasing because of $ (f(x \\plus{} \\frac 1n) \\minus{} f(x))' \\geq 0$ .. and so on ..." +} +{ + "Tag": [ + "percent" + ], + "Problem": "How much water must be evaporated from 240 gallons of a 3 percent salt solution to produce a 5 percent salt solution?", + "Solution_1": "[hide]96. This is kinda like a number sense problem. $240-.03*240*20=96$[/hide]", + "Solution_2": "is there a way to solve this question using a table or chart?" +} +{ + "Tag": [ + "blogs" + ], + "Problem": "michael jackson, the king of pop dies at age 50 :( :o . he was the best pop hands down. this is a sad day in world history", + "Solution_1": "I fondly remember making my own antigravity lean shoes and trying to learn the moonwalk. Yesterday night, I spent half an hour working on my moonwalk in the kitchen.\r\n\r\nAnyone else got any Michael Jackson memories?", + "Solution_2": "[quote=\"TZF\"]\nAnyone else got any Michael Jackson memories?[/quote]\r\nMoonwalker for the Sega mega drive (genesis). It sort of did suck, but it sure was a novelty.", + "Solution_3": "I once tried to learn the entire Thriller dance. \r\n\r\nI proved it is impossible.", + "Solution_4": "At our graduation the speaker didn't like the king very much. So he said \" Today another chapter in your guys lives ends. Michael Jackson died.\" It was pretty funny.", + "Solution_5": "There is also a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1541857#1541857]parallel discussion of this[/url] in the India subforum. I just thought it would be reasonable to link the two :).", + "Solution_6": "The king is dead, long live the king... (I never really got that saying :lol: )\r\n\r\nBillie Jean $ \\ge$ any other song", + "Solution_7": "Can someone explain to me why he changed his skin color?", + "Solution_8": "He didn't. His melanin got a freak mutation.", + "Solution_9": "What's melalin?\r\n\r\nAnyhow, nobody will ever forget michael jackson. He was the greatest.\r\n\r\nAlso the weirdest,\r\n\r\nMOST OF ALL, HE CREATED THRILLER, THE WORLD'S BEST SELLER!!!!", + "Solution_10": "[quote=\"tcjy\"]What's melalin?[/quote]\r\n\r\nits melanin\r\n\r\nand its skin pigment\r\n\r\nsee my blog for my opinion", + "Solution_11": "And now Billy Mays, the king of infomercials, died! :(", + "Solution_12": "Really? HE died? Yay, no more billy mays who yells at people in commercials while he says:\r\n\r\nBUY NOW\r\nBUY NOW!!!\r\n\r\n[b]BUY NOW!!!!!!!![/b]\r\n\r\nAnd worst of all, he advertises for things we don't need.", + "Solution_13": "[quote=\"tcjy\"]Really? HE died? Yay, no more billy mays who yells at people in commercials while he says:[/quote]\r\n\r\n\r\nDude\r\n\r\nHe may have been annoying in your eyes, but you have to RESPECT his greatness in the advertising world and at least show some grief for his death.\r\n\r\nOtherwise, when you die, people are gonna be like\r\n\r\nYay, no more tcjy who is happy when others die and doesnt respect others and never grows up.\r\n\r\nAnd worst of all, he openly admits it and seems to be proud of it", + "Solution_14": "So HE is Billy Mays.\r\n\r\nBeat It, Thriller, They Don't Care About Us, Smooth Criminal, Billy Jean, Cry, Bad, Ghost. These were his best.\r\n\r\nOf these, Beat It, Thriller, Ghost, and Billy Jean stand out.", + "Solution_15": "OMG, Billy Mays [b]DID[/b] die! :o I thought it was a joke :(\r\n\r\nGreat, everyone's dying. :noo:", + "Solution_16": "Oh no my Grandma's dieing! D:", + "Solution_17": "[quote=\"1=2\"]Beat It, Thriller, They Don't Care About Us, Smooth Criminal, Billy Jean, Cry, Bad, Ghost. These were his best. \n\nOf these, Beat It, Thriller, Ghost, and Billy Jean stand out.[/quote]\r\n\r\nHow about Heal the World, Man in the Mirror?", + "Solution_18": "[quote=\"cyberspace\"]I once tried to learn the entire Thriller dance. \n\nI proved it is impossible.[/quote]\r\n\r\nhttp://www.buzzfeed.com/mjs538/the-inmates-pay-tribue-to-michael-jackson?slideshow=most-shared\r\n :rotfl: :rotfl: :rotfl:", + "Solution_19": "[quote=\"1=2\"]So HE is Billy Mays.\n\nBeat It, Thriller, They Don't Care About Us, Smooth Criminal, Billy Jean, Cry, Bad, Ghost. These were his best.\n\nOf these, Beat It, Thriller, Ghost, and Billy Jean stand out.[/quote]\r\n\r\nHow come when everyone mentions good songs by MJ they dont mention Don't Stop 'Til You Get Enough? That was a pretty good song imo.", + "Solution_20": "Anyone that didn't manage to see the funny and entertaining side of Billy Mays' commercials really missed out.\r\n\r\nI'll miss the guy.", + "Solution_21": "But they say that he filmed so many commercials for products that arent even released yet that he will still be on TV for years.", + "Solution_22": "A dead man on TV..... OoOoOoOoO!!!", + "Solution_23": "[quote=\"jjx1\"]Can someone explain to me why he changed his skin color?[/quote]\r\n\r\n :rotfl: Ah...jjx1..you are very funny :rotfl: :rotfl:", + "Solution_24": "Didn't he have like vitiligo?\r\nIt's like an autoimmune disease that kills melanocytes.\r\n\r\nwow, he had lupus too :o", + "Solution_25": "i know i can't believe he died...\r\nlol tho it's kinda funny that now that he's dead all of a sudden like every1's buying all his albums...\r\nRIP", + "Solution_26": "Hmm..lol\r\n\r\nI'm pretty sure I can do the moon walk exactly like him. It's not that hard." +} +{ + "Tag": [ + "induction", + "function", + "trigonometry" + ], + "Problem": "prove that for every natural n \r\n$\\mid{sin(1)}\\mid + \\mid{sin(2)}\\mid + ... + \\mid{sin(3n - 1)}\\mid > \\frac{8n}{5}$", + "Solution_1": "[quote=\"violetcraze\"]prove that for every natural n \n$\\mid{sin(1)}\\mid + \\mid{sin(2)}\\mid + ... + \\mid{sin(3n - 1)}\\mid > \\frac{8n}{5}$[/quote]\r\n\r\nDoes someone have a hint, maybe induction?", + "Solution_2": "Will it be somewhat of a use if we combine the all the $sin (k)$ with their respective $sin (3n - k)$? then we can factor out $sin (\\frac{3n}{2})$. but i can't see the point of doing that", + "Solution_3": "An induction that will work: look at the function $|\\sin a| + |\\sin (a + 1)| + |\\sin (a + 2)|$ and show that it is always larger than $\\frac85$." +} +{ + "Tag": [], + "Problem": "Lon doubled a four-digit number and then subtracted 1275.\nThe result was 3573. What four-digit number did Lon double?", + "Solution_1": "Solve: $ 2x\\minus{}1275\\equal{}3573\\Rightarrow 2x\\equal{}4848\\Rightarrow x\\equal{}2424$." +} +{ + "Tag": [ + "geometry", + "rectangle", + "analytic geometry", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Consider a rectangle of the plan divided into small rectangles. It is supposed that each one of these small rectangles admits an integer side. Show that the initial rectangle admits an integer side.", + "Solution_1": "That is a very classical problem discussed more than once. The shortest solution is to integrate $e^{2\\pi i(x+y)}$ over the rectangle.", + "Solution_2": "There is an even shorter solution: just place the rectangle (parallely to the coordinate axes) into an infinite $1/2 \\times 1/2$ grid whose squares are colored in the chessboard manner. Then notice that a rectangle has (at least one) integer sidelength iff the sum of its black area equals the sum of its white area and ... that's it!" +} +{ + "Tag": [], + "Problem": "Simplify: $ \\frac{2^{12} \\minus{} 2^{10}}{2^9 \\minus{} 2^8}$", + "Solution_1": "Denominator is $ 2\\cdot2^8\\minus{}2^8\\equal{}2^8$, and numerator is $ 4 \\cdot 2^{10}\\minus{}2^{10}\\equal{}3\\cdot2^{10}$, so the answer is $ \\frac{3\\cdot2^{10}}{2^8}\\equal{}\\boxed{12}$", + "Solution_2": "I find it quicker and easier to divide both the numerator and denominator by $ 2^8$ without simplifying first, but I suppose it's just a matter of preference.", + "Solution_3": "I'm still confused.", + "Solution_4": "Which part is giving you trouble?", + "Solution_5": "[quote=CrazyAwesome88]I'm still confused.[/quote]\n\n[hide = Alternative Solution]\n$Factor$ out a $2^{10}$ in the $numerator$ \n$Factor$ out a $2^{8}$ in the $denominator$ \n\nTherefore we get,\n\n$\\frac{2^{10}*(2^{2} - 1)}{2^{8}*(2 - 1)}$\n\n$\\frac{2^{10}}{2^{8}} = 2^{2}$ \n\n$2^{2} * 3 = \\fbox{12}$\n\n\n\n[/hide]" +} +{ + "Tag": [ + "symmetry" + ], + "Problem": "Hello,\r\nThat day my teacher said that though according to Relativity length contraction occurs, there is no change in the length of a body positioned perpendicular to the direction of motion i.e. 'There is change in the length of an object only along the direction of motion'.Can anyone please give a proof of the statement?\r\nSanjith", + "Solution_1": "Well, if you want proof.\r\nThe only proof for statements in physics is observation. And, well, I do believe this is observed (or else people would go crazy and not teach us Special Relativity anymore)", + "Solution_2": "Firstly,in my opinion i want to say that the change in TIME ,LENGTH, MASS do occur(consider twin paradox).\r\nSecondly,the length of a body change because of the relativity of concurrence.When you want to measure something ,you must fix two point on it but when it goes with high velocity (C) , you must go paralell to it to measure and the relativity in concurrence occurs so the length change(what would be seen or measured if you go perpendicular to it) :D \r\nthirdly, LORRENCE believed when an electron goes with high V it looks like ellipsoide shape so its perpendicular position is changed too\r\n\r\nPS : sorry for my bad english\r\n\r\nproof:\r\nconsidered SAM is standing in the station and want to measure its length , SALLY want to do so but she sitting in the train which goes into the station. the train velocity is v\r\n\r\nSAM measure by watching the time interval when the train go though the station so:\r\n\r\n$Lo=vT$ (1)with T is the time interval when the train go though the station \r\n\r\nbut T which SAM measures is difference from SALLY so SALLY will give:\r\n\r\n$L=vTo$ (2)\r\n\r\nbut time change so with (1) and (2)\r\n$L=Lo\\sqrt{1-\\frac{v^2}{c^2}}$\r\n\r\n\r\nFROM this solution we see that the change in time lead to the change in length", + "Solution_3": "This might be what you'd call a proof that perpendicular dimensions do not change. Suppose you have a train speeding on a track. If the width of the train decreased, that would mean that its wheels would make a scratch (mark, whatever) in the ground INSIDE the tracks. But from the point of view of a passanger on a train, the track would become narrower and the wheels would clearly go OUTSIDE the tracks. This is clearly contrary to many assumptions (of symmetry, relativity and common sense) which would have to be abandoned if the perpendicular contraction was happening. And the reason why Einstein introduced (or discovered) the Lonrenz contraction was to comply with all the symmetry and relativity issues and make them compatible.", + "Solution_4": "A similar explanation was given by my teacher, a thought experiment. Consider a vertical stick inside a truck moving near a wall. Let the stick be fixed by some sort of a 'marker' with which it leaves a mark on the wall as it passes by. At ordinay speeds, it leaves a mark at a height h, say. At speeds near c, the stick would contract in length and leave a mark at a height lesser than h. This poses a paradoxical situation that violates symmetry and reference frame ideas. Hence, the vertical stick must not contract.\r\nSanjith" +} +{ + "Tag": [ + "function", + "conics", + "parabola" + ], + "Problem": "Find a function whose graph is a parabola that passes through A(1,-1), B(-1,-3), and C(3,9).", + "Solution_1": "[hide]Answer to the problem is x^2 + x -3 ... i did it quickly using a calculator... :lol: [/hide]", + "Solution_2": "[quote=\"airelemental135\"]Find a function whose graph is a parabola that passes through A(1,-1), B(-1,-3), and C(3,9).[/quote]\r\n\r\nHere is the non-calculator solution.\r\n\r\nFirst, set the function as $f(x)$ such that $f(x) = ax^2+bx+c$ for some constants $a,b,c$. Now, plug in $x = 1,-1,3$ and $f(x) = y = -1,-3,9$. From this, you should be able to determine what the $f(x)$ is." +} +{ + "Tag": [], + "Problem": "How many copies of $ \\minus{}4x\\plus{}3y$ must be added together to get $ \\minus{}24x \\plus{} 18y$?", + "Solution_1": "$ (\\minus{}24x\\plus{}18y)\\div(\\minus{}4x\\plus{}3y)\\equal{}6$\r\n\r\nanswer : 6" +} +{ + "Tag": [ + "floor function", + "linear algebra", + "linear algebra unsolved" + ], + "Problem": "Here is a cute problem:\nProve that : The maximum dimension of a commutative subalgebra of $M_n ( \\mathbb{C} )$ is equal to $\\lfloor \\frac{n^2}{4} \\rfloor + 1$.", + "Solution_1": "See these [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33640]two[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44431]topics[/url] on the same problem. The problem is already solved." +} +{ + "Tag": [ + "inequalities", + "inequalities proposed" + ], + "Problem": "For non-negative $a,b,c$ prove that\r\n\r\n$(a^3+b^3)(b^3+c^3)(c^3+a^3)\\geq 2(abc)^2\\sqrt{2(a^2+bc)(b^2+ac)(c^2+ab)}$\r\n\r\n :)", + "Solution_1": "Let $abc=1$, the problem become\r\n\\[ (a^3+b^3)(b^3+c^3)(c^3+a^3) \\ge 2\\sqrt{2(2+\\sum a^3b^3+\\sum a^3)} \\]\r\nLet $x=a^3...$ and so on. The problem become\r\n\\[ (x+y)(y+z)(z+x) \\ge 2\\sqrt{2(2+x+y+z+xy+yz+zx)} \\]\r\nNotice that\r\n\\[ \\sum xy(x+y) \\ge \\sum x+\\sum xy \\]\r\nSo, we only need to prove\r\n\\[ X+2\\ge 2\\sqrt{4+2X} \\Leftrightarrow X^2+4X+4 \\ge 4(4+2X) \\Leftrightarrow X^2-4X-12 \\ge 0 \\Leftrightarrow (X-6)(X+2) \\ge 0, \\]\r\nWhich is obviously true because $X \\ge 6$.\r\n\r\nI solved it just in a minute so i don't check it carefuly. I hope it's true.", + "Solution_2": "yes , I have checked, and it is very nice hungkhtn :D \r\n\r\nI create this inequality using Holder and AM-GM , i.e\r\n\r\n$E(a,b,c)=(a^3+b^3)(a^3+c^3)(b^3+c^3)\\\\=\\sqrt{(a^6+a^3b^3+a^3c^3+b^3c^3)(b^3c^3+a^6+a^3b^3+a^3c^3)(b^6+b^3c^3+c^6+b^3c^3)}\\\\\\geq\\sqrt{(a^2b^3c+a^3b^2c+a^2bc^3+ab^2c^3)^3}\\\\\\geq \\sqrt{8(abc)^3\\sqrt{(ab)^3}(c^2+ab)^3}$\r\n\r\nwhereby last inequality is AM-GM\r\n\r\nSimilarly we can show\r\n\r\n$E(a,b,c)\\geq\\sqrt{8(abc)^3\\sqrt{(ac)^3}(b^2+ac)^3}$\r\n$E(a,b,c)\\geq \\sqrt{8(abc)^3\\sqrt{(bc)^3}(a^2+bc)^3}$", + "Solution_3": "I found this nice solution:\r\n$\\prod (a^3+b^3)=\\prod (a+b)(a^2-ab+b^2) \\geq abc \\prod (a^2+b^2)\\geq abc \\prod (a^2+bc)\\geq 2abc\\sqrt{2\\prod (a^2+bc)}$" +} +{ + "Tag": [ + "calculus", + "derivative", + "function", + "trigonometry", + "symmetry", + "geometry", + "similar triangles" + ], + "Problem": "A fence 8 fr tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?\r\n\r\ncould someone please help me with this...i have seen this kind of problem and have NEVER been able to solve it.\r\nI normally start off w/ similar triangles but then when I take the derivative I end up with an ugly function whos root I am unable to find.", + "Solution_1": "Let the angle the ladder makes with the ground be $\\theta$, and let the distance from the ladder to the fence on the other side of the fence be $x$, then the ladder touches the building at a height of $(4+x)\\tan\\theta$, and the length of the ladder , $L$, is $(4+x)\\sec\\theta$. Now, using similar triangles, we have $\\frac{4+x}{(4+x)\\tan\\theta} = \\frac{x}{8} \\Rightarrow \\theta = \\cot^{-1}\\frac{x}{8}$. So now, $L=(4+x)\\sec\\left(\\cot^{-1}\\frac{x}{8}\\right) = (4+x)\\cdot\\frac{\\sqrt{64+x^2}}{x}\\Rightarrow \\frac{dL}{dx} = \\frac{x^3-256}{x^2\\sqrt{x^2+64}}$. Setting this equal to zero gives $x = 256^{1/3}$. Then plug this value back into $L$ to get $L\\approx 16.6478$. (I might have made a mistake somewhere since this is an ugly answer).", + "Solution_2": "its seems as if finding $\\frac{dL}{dx}$ would take a while...is there a quick way to do it?\r\ni mean i could do it with quotent rule but, thats very time consuming.", + "Solution_3": "Try working the problem in general: the height of the fence is $a,$ and it is at a distance $b$ from the wall. What is the answer, as a function of $a$ and $b?$\r\n\r\nThe answer to this will be the same as to the \"pipe around the corner\" problem: two corridors of width $a$ and $b$ meet at a right angle. What is the longest straight pipe that can be carried around the corner while remaining horizontal?\r\n\r\nThis problem has appeared here before (you could try searching for it) but it's been a while.\r\n\r\n[hide=\"Short answer\"]$\\left(a^{2/3}+b^{2/3}\\right)^{3/2}$\n\nSo $\\left(4^{2/3}+8^{2/3}\\right)^{3/2}=4\\left(1+2^{2/3}\\right)^{3/2} \\approx 16.64775$ is the correct answer.[/hide]", + "Solution_4": "thanks.\r\n \r\nbefore Jimmy posted his solution i knew the formula but had no idea how to solve it w/o using it. I also was not able to see how it related to the ladder in 2 hallways problem, but after messing with my drawing a little :P i see the relationship.\r\n\r\nthakns for all the help.", + "Solution_5": "Most of the classic first semester calculus max/min problems deserve to be revisited in third semester calculus when you learn about Lagrange multipliers. That's because most of them can be fairly naturually set up as constrained multivariable problems, and Lagrange multipliers is a method that often gives reasonable insights into problems.\r\n\r\nHere's the problem: Minimize $L^2=x^2+y^2$ subject to the constraint that the line segment from $(x,0)$ to $(0,y)$ passes through the point $(a,b).$\r\n\r\nUsing similar triangles, we can write the constaint as the equation\r\n\r\n$(x-a)(y-b)=ab.$ (I'm going to leave it in this factored form.)\r\n\r\nMatching partial derivatives with the Lagrange multiplier gives us these two equations:\r\n\r\n$2x=\\lambda(y-b)$\r\n$2y=\\lambda(x-a).$\r\n\r\nSolve each of these for $\\frac2{\\lambda}$ and equate:\r\n\r\n$\\frac{x-a}y=\\frac{y-b}x$ or $x(x-a)=y(y-b)$\r\n\r\nSolving the constraint for $y$ in terms of $x$ gives us that $y-b=\\frac{ab}{x-a}$ and $y=\\frac{ab}{x-a}+b=\\frac{ab+bx-ab}{x-a}=\\frac{bx}{x-a}.$\r\n\r\nPut this into the prevous equation to get\r\n\r\n$x(x-a)=\\left(\\frac{ab}{x-a}\\right)\\left(\\frac{bx}{x-a}\\right).$\r\n\r\n$x(x-a)^3=(ab)(bx)$ or $(x-a)^3=ab^2.$\r\n\r\nFrom this we get that $x=a+a^{1/3}b^{2/3}.$ It follows from this that $y=b+a^{2/3}b^{1/3}.$ (Note the symmetry of the expressions).\r\n\r\nSo $x^2+y^2=\\left(a+a^{1/3}b^{2/3}\\right)^2+ \\left(b+a^{2/3}b^{1/3}\\right)^2.$\r\n\r\nExpand this out, collect terms, and we get that\r\n\r\n$x^2+y^2=\\left(a^{2/3}+b^{2/3}\\right)^3.$\r\n\r\nTake square roots, and we have our result.\r\n\r\nNote that this is the only critical point, and that as either $x\\to a^+$ or $y\\to b^+,$ the length tends to $\\infty.$ Thus we have a minimum.\r\n\r\n--\r\n\r\nReference for Lagrange multipliers: Check the index of your Stewart textbook. In the 5th edition, it should be Chapter 14.", + "Solution_6": "Once you have the right function, you can use a graphing calculator to get the answer immediately.", + "Solution_7": "[quote=\"maokid7\"]its seems as if finding $\\frac{dL}{dx}$ would take a while...is there a quick way to do it?\ni mean i could do it with quotent rule but, thats very time consuming.[/quote]\r\n\r\nI remember that I posted my solution for general case.which Kent Merry field has explained. I'm sure that blahblahblah solved in using Holder.As soon as I find my post, I will post it.", + "Solution_8": "At last I found it! Here is \r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24520]www.mathlinks.ro/Forum/viewtopic.php?t=24520[/url]", + "Solution_9": "Thanks for finding it, kunny. I knew it was out there somewhere. I seem to have posted the same argument both here and there, which is somewhat redundant." +} +{ + "Tag": [ + "geometry", + "trapezoid", + "ratio", + "parallelogram", + "angle bisector", + "AMC" + ], + "Problem": "Trapezoid $ ABCD$ has $ AD\\parallel{}BC$, $ BD \\equal{} 1$, $ \\angle DBA \\equal{} 23^{\\circ}$, and $ \\angle BDC \\equal{} 46^{\\circ}$. The ratio $ BC: AD$ is $ 9: 5$. What is $ CD$?\r\n\r\n$ \\textbf{(A)}\\ \\frac {7}{9}\\qquad \\textbf{(B)}\\ \\frac {4}{5}\\qquad \\textbf{(C)}\\ \\frac {13}{15} \\qquad \\textbf{(D)}\\ \\frac {8}{9}\\qquad \\textbf{(E)}\\ \\frac {14}{15}$", + "Solution_1": "[hide]\nWe see that $ \\angle BDA\\equal{}\\angle DBC$ since $ AD\\parallel{}BC$. Draw the angle bisector $ DE$ in triangle $ BDC$. We see that triangles $ BAD$ and $ DEB$ are congruent, as they have two equal angles and share a side. Thus $ BE\\equal{}AD$. So $ BE: BC\\equal{}5: 9$, and thus $ BE: EC\\equal{}5: 4$. From the angle bisector theorem $ BE: EC\\equal{}BD: CD$, so $ CD\\equal{}\\frac{4}{5}\\equal{}B$.\n[/hide]", + "Solution_2": "[hide=\"Solution\"]Have the angle bisector of $ \\angle BDC$ meet $ BC$ at $ X$. Notice that $ ABXD$ is a parallelogram, so $ BX \\equal{} AD$ and the ratio $ XC: BX$ is $ 4: 5$. Then by the angle bisector theorem, $ CD: BD \\equal{} XC: BX$ so $ CD \\equal{} \\frac{4\\cdot1}{5} \\equal{} \\frac{4}{5}$.[/hide]" +} +{ + "Tag": [ + "geometry", + "geometric transformation", + "reflection", + "calculus", + "trigonometry", + "integration", + "analytic geometry" + ], + "Problem": "A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $ R$ be the region outside the hexagon, and let $ S\\equal{}\\{\\frac{1}{z}|z\\in R\\}$. Then the area of $ S$ has the form $ a\\pi\\plus{}\\sqrt{b}$, where $ a$ and $ b$ are positive integers. Find $ a\\plus{}b$.", + "Solution_1": "[hide=\"Hidden by mod\"]We first find the side length of the hexagon:\nLoC gives \n$ 1 \\equal{} 2x^2 \\plus{} x^2 \\equal{} 3x^2 \\\\\nx \\equal{} \\frac {\\sqrt {3}}{3}$\n\nThis makes it easy to find the two vertices with positive real parts (the ones I found easiest to work with), so we have $ v \\equal{} \\frac {1}{2} \\pm \\frac {\\sqrt {3}i}{6}$.\n\nRealizing that $ \\frac {1}{z}$ is a mapping from a point to its inversion over the unit circle, we are now equipped to find the boundary of $ S$. First of all, our vertices map to $ \\frac {1}{\\frac {1}{2} \\pm \\frac {\\sqrt {3}i}{6}} \\equal{} \\frac {3 \\mp \\sqrt {3}i}{2}$. \n\nI'm guessing that its a relatively well known result that lines inside the unit circle map to circles with their centers on the unit circle, although I actually showed this by setting up a parametric and doing a fairly long triple-substitution process :/ That's what I ended up with anyway, if anyone wants me to post the method I will.\n\nAnyway, we know realize that the boundary of $ S$ will be made up of 6 circular arcs with common endpoints. Of course, we only have to find the area inside one of these and multiply by 6. The one we know about has its center on $ 1$ and has a point at $ 2$ giving it a radius of $ 1$. We also know it covers an arc of $ \\frac {\\pi}{3}$. Thus its area is $ \\frac {\\pi}{3}$.\n\nThe remaining portion of this section of $ S$ is made up of two triangles with vertices of $ 0,1$ and $ \\frac {3 \\pm \\sqrt {3}i}{2}$, thus their lengths are $ 1,1$ and $ \\sqrt {3}$. Since these triangles have angles of $ \\frac {\\pi}{6},\\frac {\\pi}{6}$ and $ \\frac {2\\pi}{3}$, it is easy to see that they each have a height of $ \\frac {1}{2}$ and thus an area of $ \\frac {\\sqrt {3}}{4}$, giving a combined area of $ \\frac {\\sqrt {3}}{2}$.\n\nTherefore, the value we are looking for is\n\n$ 6(\\frac {\\pi}{3} \\plus{} \\frac {\\sqrt {3}}{2}) \\equal{} 2\\pi \\plus{} 3\\sqrt {3} \\equal{} 2\\pi \\plus{} \\sqrt {27}$ and so $ a \\plus{} b \\equal{} \\fbox{29}$.[/hide]", + "Solution_2": "Yeah, that is right.\r\n\r\nI just used the fact that 1/z is really just an inversion followed by a reflection over the x-axis.\r\n\r\nThen, we have that since it is symmetric over x-axis, just use some properties of inversion to figure out the shape of the graph and hence the area.", + "Solution_3": "\"R\" is usually used to define real number.\r\nA bad defintation", + "Solution_4": "[quote=\"ultramanzzy\"]\"R\" is usually used to define real number.\nA bad defintation[/quote]\r\n\r\nThis is untrue. $ \\mathbb{R}$ is the real numbers. $ R$ is left for several other things - arbitrary rings, regions (as in here), etc. Sometimes it means the real numbers, but that's rare.\r\n\r\nHaha, turns out I misread the problem as $ \\frac{1}{\\bar{z}}$, but the region doesn't change at all.", + "Solution_5": "I know you aren't supposed to need calculus, but if you do know calculus, this problem can be done in a fairly straightforward way\r\n\r\n[hide=\"using calculus\"]\nI'm going to look only at the interval $ \\theta \\in [ \\minus{} \\frac {\\pi}{6}, \\frac {\\pi}{6}]$ (the right side of the hexagon) for sake of simplicity. Since the hexagon is symmetric about the origin, multiplying this area by $ 6$ will give the answer.\n\nNow, in that interval, note that the hexagon follows\n\n$ \\cos\\theta \\equal{} \\frac {1/2}{r}$ from a simple right triangle.\n\n$ r \\equal{} \\frac {1}{2\\cos\\theta}$\n\nNow we invert it to find $ \\frac {1}{z}$'s radius. The interval will negate, but this doesn't really change anything in our problem.\n\n$ r_{inv} \\equal{} 2\\cos\\theta$, still on the same interval.\n\nNow to find the area, we can apply an integral for area in polar coordinates\n\n$ A \\equal{} \\frac {1}{2}\\int_{\\theta_a}^{\\theta_b} r^2 d\\theta$\n\n$ A \\equal{} \\frac {1}{2}\\int_{ \\minus{} \\pi/6}^{\\pi/6} (2\\cos\\theta)^2 d\\theta$\n\n$ A \\equal{} 4 \\int_0^{\\pi/6} \\cos^2\\theta d\\theta$\n\n$ A \\equal{} 2 \\int_0^{\\pi/6} 1 \\plus{} \\cos(2\\theta) d\\theta \\equal{} 2\\left[\\theta \\plus{} \\frac {\\sin(2\\theta)}{2}\\right]^{\\pi/6}_0$\n\n$ A \\equal{} 2[\\frac {\\pi}{6} \\plus{} \\frac {\\sqrt {3}}{4}]$\n\n$ 6A \\equal{} 2\\pi \\plus{} 3\\sqrt {3}$\n\nand we're all done\n[/hide]", + "Solution_6": "The construction of the region is very nice. I used a hexagon inscribed in the unit circle first, and then dilated it at the end. Here's what happens then:\r\n\r\n- Draw a hexagon that can be inscribed in a unit circle\r\n- Split this into six equilateral triangles\r\n- Draw the circumcircles of each of these equilateral triangles\r\n- The region that is their union is a scaled version of the desired region\r\n- Multiply by 1/((1/rt3)^2) = 3 to get the area", + "Solution_7": "darn i just realized that i barely looked at this problem. I think given half an hour, i could've actually cranked it out. oh well. Theres always going to be something to complain about.", + "Solution_8": "[quote=\"kops723\"]\nI'm guessing that its a relatively well known result that lines inside the unit circle map to circles with their centers on the unit circle\n[/quote]\n\nDoesn't lines map to circles that run through the reference circle? I don't think its center necessarily has to be on the reference circle.\n\n[quote=\"kops723\"]\nAnyway, we know realize that the boundary of $ S$ will be made up of 6 circular arcs with common endpoints.[/quote]\r\n\r\nHow did you get its shape?", + "Solution_9": "[quote=\"SorcererofDM\"]\nDoesn't lines map to circles that run through the reference circle? I don't think its center necessarily has to be on the reference circle.[/quote]\nYeah I guess you're right... I wasn't familiar with the result so I guess I just assumed that would be the case. It's actually only true because the midpoint of the segment is $ \\frac{1}{2}$ a unit from the origin in this case.\n\n[quote=\"SorcererofDM\"]\n[quote=\"kops723\"]\nAnyway, we know realize that the boundary of will be made up of 6 circular arcs with common endpoints.[/quote]\nHow did you get its shape?[/quote]\n\nWell the boundary of the hexagon must map to the boundary of $ S$, and since the 6 lines all map to arcs of circles with common endpoints, and are obviously otherwise disjoint, the boundary of $ S$ must be of the form I stated.\n\n[quote=\"The Zuton Force\"]\n- The region that is their intersection is a scaled version of the desired region \n[/quote]\r\nPretty sure you meant union but oh well ;)", + "Solution_10": "[quote=\"kops723\"]\nRealizing that $ \\frac {1}{z}$ is a mapping from a point to its inversion over the unit circle, \n[/quote]\r\n\r\nCan you explain what you mean by that and how you figured that out?\r\n\r\nIf $ f(z) \\equal{} 1/z$, then $ f(a \\plus{} bi) \\equal{} \\frac {a \\minus{} b i }{a^2 \\plus{} b^2}$.", + "Solution_11": "It's not just an inversion; $ f(z)\\equal{}1/z$ inverts about the unit circle then reflects about the real axis. You're right, $ f(a\\plus{}bi)\\equal{}\\frac{a\\minus{}bi}{a^2\\plus{}b^2}\\equal{}\\frac{\\overline{z}}{|z|^2}$. Recall that inversion takes $ z$ to $ \\frac{z}{|z|^2}$, since we must have $ |z\\parallel{}z'|\\equal{}1$. So inversion and reflection across the real axis do the same thing as $ f(z)\\equal{}1/z$.\r\n\r\nThe figure happens to be symmetric about the real axis, so we can just ignore the reflection component of the transformation.\r\n\r\nAs for motivation, seeing $ 1/z$ should remind one of inversion, since in standard form inversion about a unit circle takes $ OP$ to $ 1/OP$ where $ O$ is the center of inversion.", + "Solution_12": "[quote=\"SamE\"]It's not just an inversion; $ f(z) \\equal{} 1/z$ inverts about the unit circle then reflects about the real axis.[/quote]\r\n\r\nOK. I had just never heard about \"inversion about a circle\". Reading this\r\n\r\nhttp://en.wikipedia.org/wiki/Circle_inversion\r\n\r\nfilled me in.", + "Solution_13": "how do you guys learn all this stuff about the complex plane... I mean im in Calc and in algebra 2 and precalc honors we never went that in depth. Can you reccomend a book/resource? is it in Aops V.2?", + "Solution_14": "mathemonster, try the AoPS class Intermediate Trig / Complex Numbers. It helped me a lot with complex numbers.", + "Solution_15": "[quote=worthawholebean]A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $ R$ be the region outside the hexagon, and let $ S\\equal{}\\{\\frac{1}{z}|z\\in R\\}$. Then the area of $ S$ has the form $ a\\pi\\plus{}\\sqrt{b}$, where $ a$ and $ b$ are positive integers. Find $ a\\plus{}b$.[/quote]\n\nCould someone explain to me what the $ S=\\{\\frac{1}{z}|z\\in R\\}$ means? I thought the vertical line meant divisiblity. ", + "Solution_16": "It means that z is real.", + "Solution_17": "The vertical line (|) is used to express the phrase \"where\".\n\nSo its basically saying $ S=\\frac{1}{z}$ where $z \\in R$.", + "Solution_18": "Intuitively, the diagram looks like this:\n[asy]\n /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */\nimport graph; size(5 cm); \nreal labelscalefactor = 1; /* changes label-to-point distance */\npen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = -54.715, xmax = 37.085, ymin = -22.7975, ymax = 44.095; /* image dimensions */\npen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); \n\ndraw((0,17.32050807568877)--(15,8.660254037844386)--(15,-8.660254037844386)--(0,-17.32050807568877)--(-15,-8.660254037844382)--(-15,8.660254037844389)--cycle, linewidth(1) + rvwvcq); \n\ndot((0,17.32050807568877),linewidth(1pt) + dotstyle); \nlabel(\"$A$\", (0.2975,17.8375), N * labelscalefactor); \ndot((-15,8.660254037844389),linewidth(1pt) + dotstyle); \nlabel(\"$B$\", (-14.755,9.1975), NW * labelscalefactor); \ndot((-15,-8.660254037844382),linewidth(1pt) + dotstyle); \nlabel(\"$C$\", (-14.755,-8.15), SW * labelscalefactor); \ndot((0,-17.32050807568877),linewidth(1pt) + dotstyle); \nlabel(\"$D$\", (0.2975,-16.79), S * labelscalefactor); \ndot((15,-8.660254037844386),linewidth(1pt) + dotstyle); \nlabel(\"$E$\", (15.2825,-8.15), SE * labelscalefactor); \ndot((15,8.660254037844386),linewidth(1pt) + dotstyle); \nlabel(\"$F$\", (15.2825,9.1975), NE * labelscalefactor); \ndot((0,0),linewidth(2pt) + dotstyle); \nlabel(\"$O$\", (0.2975,0.5575), N * 2 * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\nNote that the problem gives us that $BF = 1$, since $\\angle BFO = 30$ degrees, $BO = \\frac{1}{\\sqrt{3}}$. Call the region to the right of $FE$ $R_1$. Clearly, for all $z \\in R_1$, $\\text{Re} (z) \\ge \\frac{1}{2}$ or in polar coordinates $z = re^{i \\theta} = r \\cos \\theta + r i \\sin \\theta$, which implies that $r \\cos \\theta \\ge \\frac{1}{2}$. Let $S_1 = \\left\\lbrace\\frac{1}{z}|z \\in R_1 \\right\\rbrace$. Then, clearly, if $R_1 = r e^ {i \\cos \\theta}$, $\\frac{1}{z} = \\frac{1}{ r e^ {i \\theta}} = \\frac{1}{r}e^ {-i \\theta} = \\frac{1}{r} \\cos \\theta - \\frac{1}{r} \\sin \\theta i$. Hence,\n$$\\frac{\\cos \\theta}{r} = \\frac{1}{2} \\Longleftrightarrow 2\\cos \\theta = r$$It\u2019s easier to visualize things in Cartesian coordinates; note that if $z = a + bi$, then $r = \\sqrt{a^2+b^2}$ and $\\cos \\theta = \\frac{a}{\\sqrt{a^2+b^2}}$. Hence,\n$$2 \\cos \\theta = r \\Longleftrightarrow 2a = a^2 + b^2 \\Longleftrightarrow (a-1)^2+b^2 = 1$$Hence, $S_1$ is a circle centered at $(1,0)$ with radius $1$. Now, after a good deal of intuition I can\u2019t really explain (it helps to think of rotation), we get that $S$ is the union of circles centered at $\\text{cis} (\\theta)$ for $\\theta = \\frac{n\\pi}{12}$, $n = 0,2,4,6,8,10$.\n[asy]\n /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */\nimport graph; size(5cm); \nreal labelscalefactor = 0.5; /* changes label-to-point distance */\npen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = -39.19, xmax = 52.61, ymin = -31.775, ymax = 35.1175; /* image dimensions */\npen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); \n\ndraw((-15,8.660254037844389)--(0,17.32050807568877)--(15,8.660254037844386)--(15,-8.660254037844386)--(0,-17.32050807568877)--(-15,-8.660254037844382)--cycle, linewidth(1) + rvwvcq); \n /* draw figures */\ndraw(circle((10,0), 10), linewidth(1)); \ndraw(circle((5,8.660254037844386), 10), linewidth(1)); \ndraw(circle((-5,8.660254037844387), 10), linewidth(1)); \ndraw(circle((-5,-8.660254037844386), 10), linewidth(1)); \ndraw(circle((5,-8.660254037844386), 10), linewidth(1)); \ndraw(circle((-10,0), 10), linewidth(1)); \ndraw((-15,8.660254037844389)--(0,17.32050807568877), linewidth(1) + rvwvcq); \ndraw((0,17.32050807568877)--(15,8.660254037844386), linewidth(1) + rvwvcq); \ndraw((15,8.660254037844386)--(15,-8.660254037844386), linewidth(1) + rvwvcq); \ndraw((15,-8.660254037844386)--(0,-17.32050807568877), linewidth(1) + rvwvcq); \ndraw((0,-17.32050807568877)--(-15,-8.660254037844382), linewidth(1) + rvwvcq); \ndraw((-15,-8.660254037844382)--(-15,8.660254037844389), linewidth(1) + rvwvcq); \ndraw((0,0)--(5,8.660254037844389), linewidth(1) + wrwrwr); \n /* dots and labels */\ndot((-5,8.660254037844386),linewidth(1pt) + dotstyle); \nlabel(\"$Z_1$\", (-4.6975,9.1975), NE * labelscalefactor); \ndot((0,0),linewidth(4pt) + dotstyle); \nlabel(\"$Z$\", (0.2975,0.5575), NE * labelscalefactor); \ndot((-10,0),linewidth(4pt) + dotstyle); \nlabel(\"$Z_6$\", (-9.76,0.5575), NE * labelscalefactor); \ndot((5,8.660254037844389),linewidth(4pt) + dotstyle); \nlabel(\"$Z_2$\", (5.2925,9.1975), NE * labelscalefactor); \ndot((10,0),linewidth(4pt) + dotstyle); \nlabel(\"$Z_3$\", (10.2875,0.5575), NE * labelscalefactor); \ndot((5,-8.660254037844384),linewidth(4pt) + dotstyle); \nlabel(\"$Z_4$\", (5.2925,-8.15), NE * labelscalefactor); \ndot((-5,-8.660254037844382),linewidth(4pt) + dotstyle); \nlabel(\"$Z_5$\", (-4.6975,-8.15), NE * labelscalefactor); \ndot((0,17.32050807568877),linewidth(4pt) + dotstyle); \nlabel(\"$Y_1$\", (0.2975,17.8375), NE * labelscalefactor); \ndot((15,8.660254037844386),linewidth(4pt) + dotstyle); \nlabel(\"$Y_2$\", (15.2825,9.1975), NE * labelscalefactor); \ndot((15,-8.660254037844386),linewidth(4pt) + dotstyle); \nlabel(\"$Y_3$\", (15.2825,-8.15), NE * labelscalefactor); \ndot((0,-17.32050807568877),linewidth(4pt) + dotstyle); \nlabel(\"$Y_4$\", (0.2975,-16.79), NE * labelscalefactor); \ndot((-15,-8.660254037844382),linewidth(4pt) + dotstyle); \nlabel(\"$Y_5$\", (-14.755,-8.15), NE * labelscalefactor); \ndot((-15,8.660254037844389),linewidth(4pt) + dotstyle); \nlabel(\"$Y_6$\", (-14.755,9.1975), NE * labelscalefactor); \nlabel(\"$1$\", (3.4025,3.7975), NE * labelscalefactor,wrwrwr); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\nNote that the length of $Y_1Y_6 = \\sqrt{3}$, so the area of $Y_1Y_6Z$ is $\\frac{3\\sqrt{3}}{4}$. It follows that the area of $Y_1Y_2Y_3Y_4Y_5Y_6$ is $\\frac{9\\sqrt{3}}{2}$. Now, we need to find the area of the lune that passes though $Y_6Y_1$, which is simply $\\frac{\\pi}{3} - \\frac{\\sqrt{3}}{4}$. Thus, the area of the onion of the circles is $2 \\pi + \\frac{6\\sqrt{3}}{2} = 2 \\pi + 3\\sqrt{3}$, giving us a final answer of $27 + 2 = \\boxed{29}$.", + "Solution_19": "[hide=unrigorous solution]\nFirst we see that $\\left|\\frac{1}{z}\\right| = \\frac{1}{|z|}$. \n\nIf we let the vertices be $ABCDEF$ where $\\overline{AB} \\cong y=1$, $O$ be the origin, and $M$ be the midpoint of $AB$. We know $M$ maps to $(0,2)$. Mapping a point $X$ on $AB$, we know $X'$ is on $\\overline{OX}$, and $OX' = \\frac{1}{OX}$. \n\nBecause $\\frac{M'O}{X'O} = 2\\cdot XO = \\frac{XO}{MO}$, and $\\angle X'OM' = \\angle XOM$, by SAS similarity, $M'X'O \\sim XMO$, and thus $M'X'O = 90 \\Rightarrow$ the locus of all $X$ is an arc with origin $(0,1)$ from $A'$ to $B'$. \n\nWe can find a similar bound on $S$ for each of the other sides of the hexagon, which are symmetrical. Then it is just some straightforward calculations.\n[/hide]\n\n" +} +{ + "Tag": [ + "calculus", + "integration", + "geometry", + "conics", + "parabola", + "analytic geometry", + "graphing lines" + ], + "Problem": "Let $c$ be the constant number such that $c>1.$Find the least area of the figure surrounded by the line passing through the point $(1,\\ c)$ and the palabola $y=x^{2}$ on $x-y$ plane.", + "Solution_1": "[quote=\"kunny\"]Let $c$ be the constant number such that $c>1.$Find the least area of the figure surrounded by the line passing through the point $(1,\\ c)$ and the palabola $y=x^{2}$ on $x-y$ plane.[/quote]\r\nThe equation of the straight line, $L,$ passing through $(1,c)$, where $c>1,$ is given by $y = m(x-1)+c,$ where $m$ is the slope of the line $L.$\r\n\r\nLet $(a,a^{2})$ and $(b,b^{2})$ be the points of intesection of the line $L$ and the parabola $y = x^{2}.$ Also, WLOG, assume $a 0$ \r\n$\\Rightarrow b-a = |b-a| = \\sqrt{{(m-2)}^{2}+4(c-1)}$ ... ($*$)\r\n\r\nNow, the area of the figure bounded by the line $L$ and the parabola $y=x^{2}$ is given by \r\n$A = \\int_{a}^{b}\\left( m(x-1)+c-x^{2}\\right)\\,dx$\r\n$= \\frac{m}{2}{(x-1)}^{2}\\Big|_{a}^{b}+cx\\Big|_{a}^{b}-\\frac{1}{3}x^{3}\\Big|_{a}^{b}$\r\n$= \\frac{(b-a)}{6}(m^{2}-4m+4c) = \\frac{1}{6}{\\left({(m-2)}^{2}+4(c-1)\\right)}^{3/2}$ ... [using $(*)$]\r\n\r\nClearly, the minimum of $A$ occurs when $m=2.$\r\n\r\nSo, $A_{min}= \\frac{1}{6}{(4(c-1))}^{3/2}= \\frac{4}{3}{(c-1)}^{3/2},$ which is our answer." +} +{ + "Tag": [ + "geometry", + "ratio", + "calculus", + "integration", + "quadratics", + "ARML", + "continued fraction" + ], + "Problem": "Congrats to Eric for getting first two years in a row! \r\n\r\nThe rankings posted on the UGA website seemed changed a little bit (not for me, though):\r\nhttp://www.math.uga.edu/olympiad/08/08top.html\r\n\r\nIn my opinion, the problems were pretty good. I did much better than last year, but still managed to make plenty of stupid mistakes on written even after I decided skipping the later questions to check...", + "Solution_1": "Yeah, the ranks aren't accurate at all.\r\n\r\nI know I bombed the test and messed up about 5 problems I normally would have gotten.\r\n\r\nCongrats to the top 40 for getting T-shirts!", + "Solution_2": "(The rankings changed because of a scoring error. It happens.)\r\n\r\nWhat were people's favorite problems?", + "Solution_3": "It turned out that they had switched the answers on two of the early problems (5 & 6 I think). Very few of the top 10 changed very much and in the top 5 teams, the only change was moving Chamblee ahead of Vestavia. Someone let Robert know that he moved up a place and I've got a $ \\$$20 check for him. My plan is to give it to Vladimir's mom to give to Vladimir to give to Robert.\r\n\r\nCongrats to everyone. You guys (and girls) did a lot better on that test than I thought folks would. I wasn't alone in thinking that 16 on the written was going to be a REALLY, REALLY good score!\r\n\r\nTom", + "Solution_4": "One may realize that if Edward had been on A team and I had been on B team, like how the practice results said and how I thought, then Walton A would have beaten 2nd place by 264 points. One may also realize that, as the only one in Walton A who didn't place in the top 10 (legally) and get a nice certificate (that is actually mine), I am actually disappointed, which is kinda freaky since 12th place still beats a lot of people....\r\n\r\nI liked how some of the problems were incredibly similar to problems that have been in other competitions. It made a 16 actually possible for people who can't do the later problems, not to mention the other people. My favorite problem is probably either 8, 17, or 19. 8 because you could use the British Flag Theorem, 17 because it's a neat trick, and 19 because it's the only geometry question on there that I thought was fun.", + "Solution_5": "Yes, that may be true. But your place on the team was earned, not given. I seem to remember a 2nd place last week at Rockdale. And then there is how much you might have contributed to the team round. The A team scored 140 to the B teams 70. I suspect that you had a part in that.", + "Solution_6": "Ha, no actually. Henry and I were one group and Claus and Abubakar were the other. We scratched our heads on #2 while they got the answer to #1 and moved on to #3, and then we switched and scratched our heads some more on #3 while they solved #2. Heh... that was pretty bad. We didn't do anything at all.\r\n\r\nThat reminds me... the answer key said something about Pell numbers for #2. I need to find out what Pell numbers are. Anyone want to prove that what the answer key said is actually true (That every one next to 1 is a Fibonacci number and that every one next to 2 is a Pell number)? I'll try but I'm not going to succeed.", + "Solution_7": "Pell numbers are defined by the recurrence P_n = 2P_{n-1} + P_{n-2} and the initial conditions P_0 = 0 and P_1 = 1. Note that the only difference from Fibonacci numbers is the presence of the \"2\". Many similar properties hold for this sequence, like the fact that the limiting ratio between consecutive elements approaches the \"silver ratio\" 1+sqrt(2). That number is so called because of many similarities to the golden ratio, such as its nice continued fraction 1+sqrt(2) = 2 + 1/(2 + 1/(2 + ...\r\n\r\nThe Pell numbers themselves are so named because they are the denominators of the best rational approximations to sqrt(2), or equivalently the y variable in integral solutions to the Pell equation x^2 - 2y^2 = +/-1. If you take every second one, as in the Markov numbers, then they're solutions for y in x^2 - 2y^2 = -1. For example, (41,29) is a solution, so 29 is a neighbor of 2 in the Markov binary tree.", + "Solution_8": "I just happened to notice that Abubakar's last name is Abid. Thus that mean that he's that MC finalist from two years ago from Lisa Academy in Arizona who just pwns math in general? Did he move? Or is Abubakar Abid just a really generic brown name?\r\n\r\nI'm really dissatisfied by my ciphering performance. For some reason, I knew exactly how to do all of the ciphering questions within the alloted time but chose for no reason in particular to screw everything up and turn in my paper early out of blind confidence. Written round was somewhat as expected, except I shouldn't have missed the one about the ratio of the two triangle's areas... The answer to team round's number three was like \"What the f***?\"\r\n\r\nAnyways, congrats to Eric and Walton (hmm, I think I've said that before XD)!!!!!!!!!!!!!!!!!!!!!!!!\r\n\r\nPS: lol, Edward's and Allen's dad was like, \"Yeah, Walton is going to beat Northview!\" No offense, but I was like, \"Thanks for the good sportsmanship and confidence booster...\" Not to insult Edward (cough, cough, cough) or Allen, but yeah, that's exactly what he said. :huh:", + "Solution_9": "Yea, Abubakar's last name is Abid. Yes, he is that MC finalist from two years ago who just pwns math in general. He did move this year. I don't know if it's generic, but I somehow doubt it.\r\n\r\nI'm glad that it boosted your confidence, but that was more supposed to be our confidence booster. It's good that it helped you too. See how well it worked?", + "Solution_10": "I thought the test was really nice. Personally i liked the tangent circle sum, the ratio of the areas, the primitive pythagorean triples and of course problem number 25. For the ratio of the areas problem, i thought it was really difficult without prior knowledge of Gergonne point, because deriving the formula is five minutes is kinda hard. Anyway i think the multiple choice pattern does help somewhat because for the fifth powers i eliminated other answer choices based on parity and using modulo 3, without using British flags. :rotfl:", + "Solution_11": "[quote=\"befuddlers\"] For the ratio of the areas problem, i thought it was really difficult without prior knowledge of Gergonne point, because deriving the formula is five minutes is kinda hard. [/quote]\r\nWell, that problem is pretty doable provided knowing that two tangents to a circle are equal and the Menelaus Theorem (or faster, mass points), even if you don't know what Gergonne point is (at least, this is the first time I heard this name). However, I missed that one by not reading the question and assuming $ X$ was the incenter...\r\n\r\nBTW, befuddlers, who are you?", + "Solution_12": "For the Gergonne point problem : you know that the ratio of $ AD$ to $ DB$ (or whatever it is) is $ (s\\minus{}a) : (s\\minus{}b)$. This gives you the ratio of triangles $ CXB$ and $ CXA$ (sorry, I'm imagining that $ X$ is where the three cevians concur; I don't know if that's actually what it's called). You can similarly compute the relative area of triangle $ AXB$, and this gets you the answer.\r\n\r\nReally, mass points are just a shortcut for arguments that rely solely on ratios of areas of triangles. You never actually need them. I really don't see how you would use Menalaus here. Do you mean Ceva? This is also a shortcut for arguments involving ratios of areas. How do you use Ceva? That only guarantees that the cevians actually concur, which is given to you in the problem anyway.\r\n\r\nIn general, \"Pell numbers\" are the solutions to the Pell equation\r\n\\[ x^2 \\minus{} Dy^2 \\equal{} \\pm 1, \\]\r\nfor positive squarefree $ D$. The general solution is kind of nontrivial (I might post something about it a little later), and involves infinite descent. Some people are fond of turning every quadratic diophantine equation with infinite descent into Pell's equation, which might explain why the solutions might talk about it and why I didn't use it at all.\r\n\r\nHow about the team round? I enjoyed it.\r\n\r\nProblem 2:\r\nFind the cardinality of the set of ordered triples $ (a,b,c)$ of positive integers satisfying $ a\\le b \\le c \\le 100$ and\r\n\\[ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3abc . \\]\r\n\r\nProblem 3:\r\nA polyhedron has a single hole (the polyhedron is evidently not convex). Incredibly, every two distinct faces of this polyhedron share exactly one edge. How many vertices does this polyhedron have?\r\n\r\n[hide=\"Problem 2\"]\nWe use a technique called \"root-flipping\", which is a standard olympiad technique. I predict that you will see it some day on a USAMO. It is useful for quadratic diophantine equations in any number of variables.\n\nSo, suppose you have a solution $ (a,b,c)$. Consider the quadratic\n\\[ x^2 \\plus{} a^2 \\plus{} b^2 \\equal{} 3abx . \\]\nYou have one solution already, $ c$. Since the quadratic's coefficients are integers, the other solution, $ c'$, must also be an integer. (This follows from Gauss's Lemma or something.) Also, since the product of $ c$ and $ c'$ is $ a^2 \\plus{} b^2$, $ c'$ is positive.\n\nNow, is $ c'$ larger or smaller than $ c$? I bet it's smaller! So now we have a smaller solution that is some permutation on the triple $ (a,b,c')$. You could do similar things with the other variables. Now, this process must eventually terminate, or else we get infinite descent (bad). So we must have solutions (which we call \"primitive\") in which we cannot replace $ a$, $ b$, or $ c$ with any smaller integer. It is not difficult to see from Viete's relations that this is the case exactly when\n\\[ c \\le \\sqrt{a^2\\plus{}b^2} , \\]\nor, equivalently, when\n\\[ c \\le \\frac{3}{2}ab . \\]\nNow, if we have all the primitive solutions, then we can reverse our root-flipping process to get all the solutions. So now we're out for primitive solutions.\n\nUsually, the way you do primitive solutions is to use the bound that guarantees primitivity to reduce the problem to a finite case check. To this end, suppose that $ a,b,c$ are all at least 2. Then\n\\[ 4b^2 \\ge 2(a^2 \\plus{} b^2) \\ge a^2\\plus{}b^2 \\plus{} c^2 \\equal{} 3abc \\ge 3ab^2 \\ge 6b^2, \\]\na contradiction. So $ a\\equal{}1$. Since $ c \\le \\sqrt{a^2 \\plus{} b^2}$ but $ c \\ge b$, it follows that $ c \\equal{}b$. It follows that $ c^2 \\equal{} b^2$ divides $ a$, so $ (a,b,c)\\equal{} (1,1,1)$ is the only primitive solution.\n\nThis means that all the solutions can be generated from (1,1,1) by root-flipping. Here, I found a diagram with a lot of arrows kind of helpful, but I'll just go throuh it here.\n\nOne flip gives us $ (1,1,3\\cdot 1 \\cdot 1 \\minus{} 1) \\equal{} (1,1,2)$. Flipping the 2 will just get us back to an old solution, so we flip another 1 and get (1,2,5). Now we could flip either the 2 or the 1.\n\nIf we flip the 1, we get (2,5,29). Then flipping the 29 will get us back where we were before, and if we try to flip 2 or 5 the numbers get too big, so we pursue this path no further.\n\nIf we flip the 2, we get (1,5,13). If we flip the 1, it becomes too large, but we can flip the 5 to get (1,13,34). If we try to flip the 1, it gets way too large, but we can flip the 13 to get $ (1,34, 3\\cdot 1 \\cdot 34 \\minus{} 13) \\equal{} (1,34, 89)$. Now, if we flip anything, we've gone too big. So we have all the solutions we want. $ \\blacksquare$\n\nNote that once you are familiar with this technique, the most difficult part if finding the primitive solutions. However, under pressure in a competition you will likely be able to guess after a short while that (1,1,1) is indeed the only primitive solution in this case, and then it is much easier to generate the other solutions by root flipping than it is to check all the $ \\binom{1000}{3}$ cases.[/hide]\r\n\r\nI might post my solution to problem 3 some time when it's not so late.", + "Solution_13": "[quote=\"1076207\"]Yea, Abubakar's last name is Abid. Yes, he is that MC finalist from two years ago who just pwns math in general. He did move this year. I don't know if it's generic, but I somehow doubt it.\n\nI'm glad that it boosted your confidence, but that was more supposed to be our confidence booster. It's good that it helped you too. See how well it worked?[/quote]\r\n\r\nTo the first part of your post, why is it that Abubakar hasn't seriously won a competition when he's supposed to be uber-amazing? He should be pwning at all of the math competitions... :huh: \r\n\r\nAnd to the second part of your post, I see sportsmanship runs in the family. :wink: \r\n\r\nNOTE: I am not flaming Allen in any way. [color=black][size=0]Only his dad to some extent.=P[/size][/color]", + "Solution_14": "Enough! Yes, Walton and Northview are both pretty good this year. Yes, Walton likes to beat Northview just as much as Northview likes to beat Walton. Remember last year? Northview won everything up to the state tournament. Who's to say that the same thing won't happen this year? You guys are too good for this bickering. AND you will have to work together in May! (And this coming weekend!) Personally, while winning Rockdale and UGA is great, I REALLY, REALLY like the fact that we both beat Vestavia both weekends! Now I'd like to concentrate on beating TJ this weekend.\r\n\r\nSee you soon,\r\nTom\r\n\r\nPS. Don't worry about Abubakar. I suspect that he will win his share. You guys are a really strong group and (except for Eric apparently) anyone can win any week.", + "Solution_15": "hey guys\r\ni realized there is no need for gergonne point, i solved the problem using equal tangents, and so can find the ratio of the sides. After finding the ratio of the sides the raios of the areas cud be found easily.\r\n\r\ncud somebody answer this plz\r\nhow to include diagrams in our posts", + "Solution_16": "[quote=\"Boy Soprano II\"]Really, mass points are just a shortcut for arguments that rely solely on ratios of areas of triangles. You never actually need them. I really don't see how you would use Menalaus here. Do you mean Ceva? This is also a shortcut for arguments involving ratios of areas. How do you use Ceva? That only guarantees that the cevians actually concur, which is given to you in the problem anyway.\n[/quote]\r\n\r\nYeah, I realized neither Ceva/Menelaus or mass points are necessary. Anyways, if you use Menelaus on, say, triangle $ APC$ and line $ QXB$, you can get ratio $ AX: XP$ which solves the problem. I suspect Menelaus will be useful when, instead of given the ratio of sides of the triangle, you are given the ratio of the segments of cevians.", + "Solution_17": "This is absolutely phenomenal :) Congratulations to Eric again for an amazing victory! And another congratulations to Edward.. that's really amazing for a 7th grader :) I think the only person who I know who came close to that was Harrison :) But it looks like the Georgia ARML team is going to be secure for the next few years or so with Sitan, Edward, and Arvind. Good luck!\r\n\r\nhaha, it looks like you're going to have a tough time choosing ARML A and B this year, Mr. Fulton :wink:\r\n\r\nI especially liked the power round questions. Too bad there aren't that many comps in college.. I miss them :(", + "Solution_18": "Okay guys, let's get something straight. LISA Academy is in Arkansas, not Arizona! Secondly, Abubakar Abid rapes math like there's no tomorrow, so piss off. :ninja:", + "Solution_19": "You might want to notice that this thread is almost 2 years old. Lots has happened since then." +} +{ + "Tag": [ + "real analysis", + "real analysis unsolved" + ], + "Problem": "Hi!\r\n\r\nit is not very hard to prove that:\r\n\\[ \\sum_{j=1}^\\infty\\sum_{i=1}^\\infty {{(i-1)!(j-1)!}\\over{(i+j)!}}=\\zeta(2). \\]\r\nCan you compute the following:\r\n\\[ \\sum_{j=1}^\\infty\\sum_{i=1}^\\infty {{(i-1)!(j-1)!}\\over{(i+j)!}}x^{i+j}. \\]\r\n\r\nMichael", + "Solution_1": "NO IDEAS :? :?" +} +{ + "Tag": [ + "function", + "real analysis", + "real analysis theorems" + ], + "Problem": "Give an example of locally unbounded function. I was thinking something in line of Riemann function... :huh: .", + "Solution_1": "try\r\nf(x)=0 if x irrational , \r\n pq if x=p/q rational" +} +{ + "Tag": [ + "AMC", + "inequalities", + "number theory", + "Intermediate Number Theory" + ], + "Problem": "Let $ S_i$ be the set of all integers $ n$ such that $ 100i\\leq n < 100(i \\plus{} 1)$. For example, $ S_4$ is the set $ {400,401,402,\\ldots,499}$. How many of the sets $ S_0, S_1, S_2, \\ldots, S_{999}$ do not contain a perfect square?", + "Solution_1": "Use the fact that $ (n \\plus{} 1)^2 \\minus{} n^2 \\equal{} 2n \\plus{} 1$\r\n\r\nAfter $ 50^2 \\equal{} 2500$, all squares fall into different sets.\r\n\r\nSince $ \\lfloor100000^\\frac {1}{2}\\rfloor$=316, we have:\r\n\r\n$ 1000 \\minus{} (266 \\plus{} 26) \\equal{} 708$ sets without squares.", + "Solution_2": "I don't agree.\r\nwhen i=998, there's one square,316^2.\r\nthe same with i=996,994,992....102\r\nOnly i=101,103,104 .. satisfy.", + "Solution_3": "It's not every other set that has a square. \r\n\r\nI got 708 as well. :)", + "Solution_4": "100i<=n<100(i+1)\r\nWHEN i=102, 10200<=n<10300, n=10201=101^2,\r\nwhen i=104, 10400<=n<10500,n=10404=102^2,\r\nwhen i=106,10600<=n<10700, n=10609=103^2", + "Solution_5": "Yeah, but it doesn't continue that way. The spacing between squares keeps increasing.", + "Solution_6": "I think i got it wrong.\r\nwhen i=606, there's no perfect square.", + "Solution_7": "From $ 50^2 \\equal{} 2500$ on, the spacing for each $ 10$ perfect squares increases by $ 2$, so from $ 50^2$ to $ 60^2$, there is one set without a perfect square, from $ 60^2$ to $ 70^2$ there are 3 such sets, and on and on until $ 316$. I got 703 though, so maybe this pattern doesn't hold :(", + "Solution_8": "The difference between consecutive squares is\r\n\\[ (x \\plus{} 1)^2 \\minus{} x^2 \\equal{} 2x \\plus{} 1,\r\n\\]which means that all squares above $ 50^2 \\equal{} 2500$ are more than 100 apart.\r\nThen the first 25 sets ($ S_1,\\cdots S_{25}$) each have at least one perfect square. Also, since $ 316^2 < 100000 < 317^2$, there are $ 316 \\minus{} 50 \\equal{} 266$ other sets after $ S_{25}$ that have a square. Then there are $ 1000 \\minus{} 266 \\minus{} 26 \\equal{} 708$ without a perfect square.", + "Solution_9": "I got 292 (1000-708) :ewpu:", + "Solution_10": "Noooo! I subtracted from 999 instead of 1000, to get 707 :wallbash:", + "Solution_11": "To keep up the string of dumb mistakes, $ 316^2 \\equal{} 100856$, and so I got 709. :(", + "Solution_12": "Isnt the answer 709?\r\n\r\nThe first 26 sets have perfect squares; S0 through S25.\r\nDid you guys forget to count S0?\r\nAnd then 1000 - (265+26) = 709?", + "Solution_13": "[quote=\"phoenixstrike\"]Isnt the answer 709?\n\nThe first 26 sets have perfect squares; S0 through S25.\nDid you guys forget to count S0?\nAnd then 1000 - (265+26) = 709?[/quote]\r\n\r\nS0 to S25 is 26\r\nthen the squares 51^2 to 316^2 each correspond to an S. That's 26[b]6[/b]. \r\n1000 - (266 + 26) = 708.", + "Solution_14": "[quote=\"nr1337\"][quote=\"phoenixstrike\"]Isnt the answer 709?\n\nThe first 26 sets have perfect squares; S0 through S25.\nDid you guys forget to count S0?\nAnd then 1000 - (265+26) = 709?[/quote]\n\nS0 to S25 is 26\nthen the squares 51^2 to 316^2 each correspond to an S. That's 26[b]6[/b]. \n1000 - (266 + 26) = 708.[/quote]\r\nWell I just did it this way:\r\nThere are $ (999\\minus{}26\\plus{}1)$ $ S$'s from $ S_{26}$ to $ S_{999}$, and there are $ (316\\minus{}51\\plus{}1)$ squares from $ 51$ to $ 316$, and each square corresponds to an $ S$, so the answer is:\r\n$ (999\\minus{}26\\plus{}1)\\minus{}(316\\minus{}51\\plus{}1)\\equal{}708$", + "Solution_15": "Don't look at the question as \"how many sets have a square in them\". Look at the question as \"how many squares have a set\". Then it's easy to see that there are multiple squares in each set for the first 26 sets. Then all integers from 51-316 will have a set, so the answer is 1000 - 26 - (316-51+1) = 708 sets without a square.", + "Solution_16": "Can someone explain to me how you know there are 50 sets with perfect squares after set 25?", + "Solution_17": "We use complementary counting. After $50^2$, all squares are in different sets of $100$, from $51^2$ to $316^2$, for $266$ sets. For $50^2$ or below, all sets work, giving an additional $26$ sets. The total is $266+26 = 792$, for an answer of $1000-792=\\boxed{208}$.", + "Solution_18": "[quote=OlympusHero]We use complementary counting. After $50^2$, all squares are in different sets of $100$, from $51^2$ to $316^2$, for $266$ sets. For $50^2$ or below, all sets work, giving an additional $26$ sets. The total is $266+26 = 792$, for an answer of $1000-792=\\boxed{208}$.[/quote]\n\nthis thread is from 2008 :| ", + "Solution_19": "[quote=Notaparrot][quote=OlympusHero]We use complementary counting. After $50^2$, all squares are in different sets of $100$, from $51^2$ to $316^2$, for $266$ sets. For $50^2$ or below, all sets work, giving an additional $26$ sets. The total is $266+26 = 792$, for an answer of $1000-792=\\boxed{208}$.[/quote]\n\nthis thread is from 2008 :|[/quote]\n\nI think you're allowed to post solutions to old contest problems", + "Solution_20": "[quote=brainfertilzer][quote=Notaparrot][quote=OlympusHero]We use complementary counting. After $50^2$, all squares are in different sets of $100$, from $51^2$ to $316^2$, for $266$ sets. For $50^2$ or below, all sets work, giving an additional $26$ sets. The total is $266+26 = 792$, for an answer of $1000-792=\\boxed{208}$.[/quote]\n\nthis thread is from 2008 :|[/quote]\n\nI think you're allowed to post solutions to old contest problems[/quote]\n\nyes, even if it is similar to other solutions", + "Solution_21": "[insert off topic post about how this thread was made before me]$ $", + "Solution_22": "[quote=asdf334][insert off topic post about how this thread was made before me]$ $[/quote]\n\n", + "Solution_23": "[quote=asdf334][insert off topic post about how this thread was made before me]$ $[/quote]\n\nloool fr", + "Solution_24": "dum problem, it is easy to see that no set has a square. there are only squiggles, ovals, and diamonds." +} +{ + "Tag": [ + "modular arithmetic", + "number theory solved", + "number theory" + ], + "Problem": "i have a question :\r\n\r\nhow to evaluate the last 6 digits of 5^131313?\r\n\r\ni want to use the Euler's Theorem, but gcd(5,10^6) is not 1, so what theory shall i use to evaluate the results?\r\n \r\nthank you very much!", + "Solution_1": "Let $5^{131313} = 10^6 q + r$, where $0 \\le r < 10^6$. Then $5^{131313} = 2^6 5^6 q + r$, so $r$ is divisible by $5^6$. Let $r = 5^6 s$. \r\n\r\nThen $5^{131313} = 2^6 5^6 q + 5^6 s$, or $5^{131307} = 2^6 q + s$. So, we are reduced to finding $5^{131307} \\pmod{2^6}$. Try taking it from there.", + "Solution_2": "Generally when computing large residues it is helpful to introduce the Chinese Remainder Theorem. In this case you are working with modulo $10^6 = 1000000$, which is equivalent to working simultaneously in modulos $5^6 = 15625$ and $2^6 = 64$. You will usually find that Euler's result applies or the residue is trivially zero." +} +{ + "Tag": [ + "ceiling function", + "combinatorics proposed", + "combinatorics" + ], + "Problem": "Let $E$ be a family of subsets of $\\{1,2,\\ldots,n\\}$ with the property that for each $A\\subset \\{1,2,\\ldots,n\\}$ there exist $B\\in F$ such that $\\frac{n-d}2\\leq |A \\bigtriangleup B| \\leq \\frac{n+d}2$. (where $A \\bigtriangleup B = (A\\setminus B) \\cup (B\\setminus A)$ is the symmetric difference). Denote by $f(n,d)$ the minimum cardinality of such a family.\r\na)\tProve that if $n$ is even then $f(n,0)\\leq n$.\r\nb)\tProve that if $n-d$ is even then $f(n,d)\\leq \\lceil \\frac n{d+1}\\rceil$.\r\nc)\tProve that if $n$ is even then $f(n,0) = n$", + "Solution_1": "A) Let $E$ be the family of sets $a_{i}= \\{1, 2, \\cdots, i\\}, i = 1, 2, \\cdots, n$. Given any set $S$ with $|S| = k$. Then $|S \\bigtriangleup a_{1}| = k \\pm 1, |S \\bigtriangleup a_{i+1}| = |S \\bigtriangleup a_{i}| \\pm 1$, and $|S \\bigtriangleup a_{n}| = n-k$. So either one of either $|S \\bigtriangleup a_{1}|$ or $|S \\bigtriangleup a_{n}|$ is equal to $\\frac{n}{2}$, or they lie on opposite sides of $\\frac{n}{2}$. In the latter case, one of the intermediate $a_{i}$ must be $\\frac{n}{2}$.\r\n\r\nB) Let $E$ be the family of sets ${a_{i}= \\{1, 2, \\cdots,1+(d-1) i}$ such that $0 \\leq i < \\lceil \\frac{n}{d+1}\\rceil = j$. Given any set $S$ with $|S| = k$. Then $|S \\bigtriangleup a_{0}| = k \\pm 1, | |S \\bigtriangleup a_{i+1}|-|S \\bigtriangleup a_{i}| | \\leq d+1$, and $| |S \\bigtriangleup a_{j}|-(n-k) | \\leq d$. Then both $|S \\bigtriangleup a_{0}|$ and $|S \\bigtriangleup a_{0}|$ cannot be less than $\\frac{n-d}{2}$, since their sum is at least $n-d-1$. Similarly, they cannot both be more than $\\frac{n-d}{2}$. So either one of them lies within the given range, or they lie on opposite sides. So one of the intermediate $a_{i}$ must lie within the range, as $d+1$ is not long enough to cross it." +} +{ + "Tag": [ + "algebra", + "function", + "domain" + ], + "Problem": "I was doing Piece-Wise Defined graphs in class and have a difficult time solving for domain. Could someone share an example of solving for domain of a random piece-wise graph, with two equations and only one stipulation per equation?", + "Solution_1": "wut is a Piece-Wise Defined Graph and whatis its domain. sry, but i've this first time", + "Solution_2": "[quote=\"analytic\"]wut is a Piece-Wise Defined Graph and whatis its domain. sry, but i've this first time[/quote]\n\nA piece-wise function is a function that is composed of two or more different functions in different domains, for example $ f(x)\\equal{}\\begin{cases}x\\plus{}1&\\text{if } x\\geq 0\\\\\\minus{}x&\\text{if } x<0\\end{cases}$.\n\n[quote=\"Slippy\"]I was doing Piece-Wise Defined graphs in class and have a difficult time solving for domain. Could someone share an example of solving for domain of a random piece-wise graph, with two equations and only one stipulation per equation?[/quote]\r\n\r\nAnd Slippy, I don't understand your question. Sorry. What do you mean by stipulation? Also, a domain is given inside a piecewise function.", + "Solution_3": "try to graph a piece-wise defined graph and you'll understand without any help.", + "Solution_4": "The domain is the combined domain of the individual parts with the restrictions.\r\nSo for the example function, since both parts have unbounded domain, we would say the domain would be $ x \\ge 0\\ \\text{and}\\ x < 0$, so the domain is all real numbers." +} +{ + "Tag": [ + "inequalities proposed", + "inequalities" + ], + "Problem": "Let $ a,b,c$ be positive reals such that $ abc\\geq 1$. Prove that\r\n\r\n$ \\frac {a \\plus{} 1}{a^2 \\plus{} a \\plus{} 1} \\plus{} \\frac {b \\plus{} 1}{b^2 \\plus{} b \\plus{} 1} \\plus{} \\frac {c \\plus{} 1}{c^2 \\plus{} c \\plus{} 1}\\leq 2$.", + "Solution_1": "[quote=\"discredit\"]Let $ a,b,c$ be positive reals such that $ abc\\geq 1$. Prove that\n\n$ \\frac {a \\plus{} 1}{a^2 \\plus{} a \\plus{} 1} \\plus{} \\frac {b \\plus{} 1}{b^2 \\plus{} b \\plus{} 1} \\plus{} \\frac {c \\plus{} 1}{c^2 \\plus{} c \\plus{} 1}\\leq 2$.[/quote]\r\nIt is equivalent to $ (abc\\minus{}1)\\sum_{cyc}\\left(\\frac{1}{3}\\plus{}a\\plus{}ab\\plus{}\\frac{2}{3}abc\\right)\\plus{}\\frac{1}{2}\\sum_{cyc}c^2(a\\minus{}b)^2\\geq0.$", + "Solution_2": "Wow, arqady. How do you get that?\r\n\r\nAnd any simpler solutions?", + "Solution_3": "fisrt,you should solve this problem:\r\nx^2/(x^2+y^2+xy) + y^2/(y^2+z^2+yz) + z^2/(z^2+x^2+xz) >=1\r\nthen I 'll post the solutions for your problem! :D", + "Solution_4": "[quote]Math pro wrote:\nfirst,you should solve this problem: \nx^2/(x^2+y^2+xy) + y^2/(y^2+z^2+yz) + z^2/(z^2+x^2+xz) >=1 \nthen I 'll post the solutions for your problem! [/quote]\r\n\r\nWhat on the earth does connect these two problems??????", + "Solution_5": "[quote]Math pro wrote: \nfirst,you should solve this problem: \nx^2/(x^2+y^2+xy) + y^2/(y^2+z^2+yz) + z^2/(z^2+x^2+xz) >=1 \nthen I 'll post the solutions for your problem! [/quote]\r\n\r\nHere's your solution Math pro. But I feel it's not alright!\r\n\r\nx^2/(x^2+y^2+xy) + y^2/(y^2+z^2+yz) + z^2/(z^2+x^2+xz) >= [b]x^2/(x+y)^2+y^2/(y+z)^2+z^2/(z+x)^2 >= x^2/(y+z)^2+y^2/(z+x)^2+z^2/(x+y)^2[/b]\r\nor, x^2/(y+z)^2+y^2/(z+x)^2+z^2/(x+y)^2 >= x/(y+z)+y/(z+x)+z/(x+y) >= 1.5 > 1........Done!\r\n[This is not my solution, it has been copied line by line from a book. I myself have some query in the highlighted step. If you could make out please inform me]\r\n :D", + "Solution_6": "I think there's no connection, Math Pro. Even if there is, if you don't want to post the solution, please don't do this. Create a separate topic :wink:", + "Solution_7": "[quote=\"discredit\"]How do you get that?\n[/quote]\r\nThe expanding gives it easy enough. :wink:", + "Solution_8": "[quote=\"Sunkern_sunflora\"][quote]Math pro wrote: \nfirst,you should solve this problem: \nx^2/(x^2+y^2+xy) + y^2/(y^2+z^2+yz) + z^2/(z^2+x^2+xz) >=1 \nthen I 'll post the solutions for your problem! [/quote]\n\nHere's your solution Math pro. But I feel it's not alright!\n\nx^2/(x^2+y^2+xy) + y^2/(y^2+z^2+yz) + z^2/(z^2+x^2+xz) >= [b]x^2/(x+y)^2+y^2/(y+z)^2+z^2/(z+x)^2 >= x^2/(y+z)^2+y^2/(z+x)^2+z^2/(x+y)^2[/b]\nor, x^2/(y+z)^2+y^2/(z+x)^2+z^2/(x+y)^2 >= x/(y+z)+y/(z+x)+z/(x+y) >= 1.5 > 1........Done!\n[This is not my solution, it has been copied line by line from a book. I myself have some query in the highlighted step. If you could make out please inform me]\n :D[/quote]\r\nare you sure? :blush:", + "Solution_9": "set up a=k.x/y b=k.y/z c=k.z/x\r\nthen use my problem! :blush:", + "Solution_10": "[quote]Math pro wrote:\nAre you sure? :blush: [/quote]\r\nI told you that I myself is a bit confused with the solution. But my intution says it is correct, although I didn't get the step still now. After all, it is from a book! :wink:", + "Solution_11": "this is my solution for my problem:\r\nset up y/x=ab/c^2, y/z=bc/a^2, z/x=ca/b^2\r\nanh then use cauchy-Schwarz\r\nI'm sure that it is not hard for you! :blush:" +} +{ + "Tag": [], + "Problem": "If $ C\\equal{}\\frac{5}{9}(F\\minus{}32)$, what is $ F$ when $ C\\equal{}20$?", + "Solution_1": "$ C\\equal{}\\frac{5}{9}(F\\minus{}32) \\iff F\\equal{}\\frac{9}{5}C\\plus{}32 \\equal{} 36\\plus{}32 \\equal{} \\boxed{68}$." +} +{ + "Tag": [ + "limit", + "integration", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Find \\[ \\lim_{n\\to\\infty} \\int_{0}^{n} (1+\\frac{x}{n})^{n} e^{-2x} dx. \\]", + "Solution_1": "$\\int_{0}^{n} (1+\\frac{x}{n})^{n} e^{-2x} dx = \\int_0^\\infty (1+\\frac xn)^n e^{-2x} \\chi_{[0,n]}(x)dx$\r\n\r\n$|(1+\\frac xn)^n e^{-2x} \\chi_{[0,n]}(x)| \\leq (1+\\frac xn)^n e^{-2x} \\leq e^x e^{-2x} = e^{-x} \\in L^1([0,\\infty))$.\r\n\r\nBy Lebesgue's Dominated Convergence Theorem,\r\n\\[ \\lim_{n\\to\\infty}\\int_{0}^{n} (1+\\frac{x}{n})^{n} e^{-2x} dx = \\int_0^\\infty \\lim_{n\\to\\infty} (1+\\frac xn)^n e^{-2x} \\chi_{[0,n]}(x)dx = \\int_0^\\infty e^{-x}dx = 1 \\]", + "Solution_2": "Can you show, that $f_n \\to f$ by measure?", + "Solution_3": "Hm? Here the convergence is pointwise...", + "Solution_4": "Yes, i see... i thought that there must be convergence by measure, but then i looked to book, and find out that therem is right for pointwise converegence and for convergence by measure...So you are right! :)" +} +{ + "Tag": [], + "Problem": "Is it allowed to give multiple proofs for a problem, so if one proof is wrong and the other is correct then you'll still get full credit? Will you get marked down in any way for this?\r\n\r\nAlso, if you prove an extension to a problem (e.g. the problem asks for a specific case but you prove the general case), can you provide an additional solution for the specific case?", + "Solution_1": "Look down maybe 20 topics to the one with the title: \"submitting 2 solutions\" for a response." +} +{ + "Tag": [ + "linear algebra" + ], + "Problem": "$A,b\\in M_n(R)$ \r\n$y>x>0$ such that $xBA-yAB=(y-x)I_n$\r\n\r\nProve that $det(AB-BA)=0$", + "Solution_1": "From the problem condition we can easily get $x(BA+I_n)=y(AB+I_n)$ (1) and using the fact that the square matrices $AB$ and $BA$ have the same set of eigenvalues we conclude that $det(BA+I_n)=det(AB+I_n)$ which implies according to (1) that $det(BA+I_n)=det(AB+I_n)=0$ (2). Also from the problem condition we can get $y(BA-AB)=(BA+I_n)(y-x)$ from where according to the aforementioned facts the conclusion follows", + "Solution_2": ":first:" +} +{ + "Tag": [ + "inequalities proposed", + "inequalities" + ], + "Problem": "Let $ a,b,c$ be the positive integers satifying $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 2(ab \\plus{} bc \\plus{} ca)$.Find the maximum value of $ M \\equal{} \\frac {a^4 \\plus{} b^4 \\plus{} c^4}{(a \\plus{} b \\plus{} c)^4}$", + "Solution_1": "I found that $ M_{max}\\equal{}\\frac{7}{24}$ is it true?", + "Solution_2": "[quote=\"SUPERMAN2\"]Let $ a,b,c$ be the positive integers satifying $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 2(ab \\plus{} bc \\plus{} ca)$.Find the maximum value of $ M \\equal{} \\frac {a^4 \\plus{} b^4 \\plus{} c^4}{(a \\plus{} b \\plus{} c)^4}$[/quote]\r\n\r\nLet assume $ a\\plus{}b\\plus{}c \\equal{}1$;\r\n\r\nby $ pqr$ we have : $ p\\equal{}1$ , $ q\\equal{}\\frac{1}{2}$ and $ 0 \\le r \\le \\frac{1}{54}$;\r\n\r\n$ M \\equal{} \\frac {a^4 \\plus{} b^4 \\plus{} c^4}{(a \\plus{} b \\plus{} c)^4}\\equal{} \\frac{\\minus{}p^{4}\\plus{}8p^2q^2\\plus{}2q^4}{9p^4}\\plus{}4\\frac{r}{p^3}\\equal{}\\frac{1}{8}\\plus{}4r \\le \\frac{43}{216}$;\r\n\r\nIquality holdes for $ a\\equal{}b\\equal{}\\frac{1}{6}$ and $ c\\equal{}\\frac{2}{3}$;", + "Solution_3": "[quote=\"Sergic Primazon\"][quote=\"SUPERMAN2\"]Let $ a,b,c$ be the positive integers satifying $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 2(ab \\plus{} bc \\plus{} ca)$.Find the maximum value of $ M \\equal{} \\frac {a^4 \\plus{} b^4 \\plus{} c^4}{(a \\plus{} b \\plus{} c)^4}$[/quote]\n\nLet assume $ a \\plus{} b \\plus{} c \\equal{} 1$;\n\nby $ pqr$ we have : $ p \\equal{} 1$ , $ q \\equal{} \\frac {1}{2}$ and $ 0 \\le r \\le \\frac {1}{54}$;\n\n$ M \\equal{} \\frac {a^4 \\plus{} b^4 \\plus{} c^4}{(a \\plus{} b \\plus{} c)^4} \\equal{} \\frac { \\minus{} p^{4} \\plus{} 8p^2q^2 \\plus{} 2q^4}{9p^4} \\plus{} 4\\frac {r}{p^3} \\equal{} \\frac {1}{8} \\plus{} 4r \\le \\frac {43}{216}$;\n\nIquality holdes for $ a \\equal{} b \\equal{} \\frac {1}{6}$ and $ c \\equal{} \\frac {2}{3}$;[/quote]\r\n\r\nArn't $ a,b,c$ positive integers?", + "Solution_4": "Oh I'm sorry.$ a,b,c$ are positive real numbers.\r\nEDIT:Let $ a,b,c$\u2265$ 0$ satisfying $ a^2\\plus{}b^2\\plus{}c^2\\equal{}2(ab\\plus{}bc\\plus{}ca)$.Find the maximum value of $ M\\equal{}\\frac{a^4\\plus{}b^4\\plus{}c^4}{(a\\plus{}b\\plus{}c)^4}$" +} +{ + "Tag": [ + "AwesomeMath", + "summer program", + "Stanford", + "college", + "blogs", + "ARML", + "HMMT" + ], + "Problem": "Hey...now that Awesomemath 09 is over....it feels really weird...so everyone please reply if you were @ cali!!!!!", + "Solution_1": "Hmm, I was there...", + "Solution_2": "So was I. Needs more text before submitting.", + "Solution_3": "I know who the three above pplz are. Yay.\r\n\r\nThe question is who knows me?", + "Solution_4": "Albino. Is that enough proof?", + "Solution_5": "You want to ride your biCYCLE! You want to ride your bike... You want to ride your biCYCLE! You want to ride it where you like...", + "Solution_6": "Ravicycle!", + "Solution_7": "Spawn of the Devil. Even more proof for you.", + "Solution_8": "Monkey Face. LOL.", + "Solution_9": "i dont know if im doing this right. FIRST POST. yah, i was at ucsc. I MISS ALL YOU AMSP PEOPLE ]: i dont think im doing this post thing right.", + "Solution_10": "[quote=\"AIME15\"]Monkey Face. LOL.[/quote]\r\nUmmm.... haven't heard that one before...", + "Solution_11": "Seriously? It started on Day 3...and considering the amount of time you spent in our room...", + "Solution_12": "wait, who's MonkeyFace?\r\nDerek?\r\nCause if it is, then i totally second that :P\r\n[in a good way]\r\nThere were a lot of monkey faces apparently.", + "Solution_13": "No, not Derek. It's someone who's posted in this thread already :D\r\n\r\nBTW, who are you?", + "Solution_14": "someone who took a pic of you humping that pole outside of parkman\r\non the last day of camp. yah, when we were signing yrbook i think.\r\nlol [:", + "Solution_15": "is this the biggest post-thingy ever in the history of awesomemath posts?(i mean overall)", + "Solution_16": "yups. biggest post. cause amsp 09 santa cruz ROCKKKKKKKKKED.\r\nman im hungry. O_O", + "Solution_17": "I miss you guys so much. Wish we could all just hang out, hopefully we'll all go to MIT and chill everyday. Cus that would be awesome. Makes me tear a bit whenever I think about camp and how much it has changed my life. Like, I have become someone because of camp. Marc has made me really get involved with community and service and I no longer suck badly at math...", + "Solution_18": "i miss camp...", + "Solution_19": "Who here has Skype other than me?", + "Solution_20": "meee. what's your skype?", + "Solution_21": "JESSICA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!wow. hows sc?????????does it suck??wow...its rlyrly weird now that ur not a floridian...guess its just me and eli now...:(YOU HAD TO LEAVE ME WITH THE AWKWARD PERSON!!!!!!!!jk eli if u read this...so any...HEY EVERYONE!!!!!i miss camp too...", + "Solution_22": "oneequalstwo is mah skype. :D", + "Solution_23": "this is like spamming \r\n\r\njust like the missouri forum.", + "Solution_24": "spam is awesome. SPAM SPAM SPAM\r\n\r\nomg jess, hows south carolinaaa?\r\nit doesnt affect ME much but stilllll\r\nis ur new school okkkay?\r\nare there any koreannns? there are like 4 at our school O_O\r\nwell yah, pretty much.\r\n\r\nshri, i dont think u became a person yet. jkjk [:\r\ni miss campppppp are u guys going next yr?", + "Solution_25": "im DEFINITELY going bak nxt yr!!!!!!!!!!!!!im excited already. did u kno that t2 already uploaded the awesomemath site for 2010???wow....^^", + "Solution_26": "Helen wtf are you talking about, there are a million koreans at our school. I can name like 20 off the top of my head.\r\n\r\nI think I will apply to be a counselor for CA next summer :)", + "Solution_27": "POSTZ YOURZ ZKYPEZ AND AIMZ!!!\r\n\r\nAIM: nuts for the It\r\nSkype: oneequalstwo", + "Solution_28": "[quote=\"serialk11r\"]Helen wtf are you talking about, there are a million koreans at our school. I can name like 20 off the top of my head.\n\nI think I will apply to be a counselor for CA next summer :)[/quote]\r\n\r\nserial this is no place for racism\r\n\r\nDONT BE RACIST\r\n\r\nYOU GUYS GOT REPORTED.", + "Solution_29": "Merely saying that there are many Koreans at a school is not racism." +} +{ + "Tag": [ + "group theory", + "abstract algebra", + "superior algebra", + "superior algebra solved" + ], + "Problem": "Prove that a finite group cannot be written as the union of the conjugates of a proper subgroup.", + "Solution_1": "Let $U$ be a proper subgroup of the finite group $G$.\r\nWe have $y^{-1}x \\in U \\Rightarrow xUx^{-1}=yUy^{-1}$ and for every $y \\in G$ there are the $|U|$ elements $x \\in yU$ so that $y^{-1}x \\in U$. Now let be $I$ a subset of $G$ representing these cosets. We get $|I|=\\frac{|G|}{|U|}>1$ and with that (using that the neutral element $e$ is in every conjugate of $U$):\r\n$| \\bigcup_{x \\in G} xUx^{-1}| = | \\bigcup_{x \\in I} xUx^{-1}| \\leq |I| \\cdot (|U|-1)+1=|G|-|I|+1<|G|$, so $G$ is not union of the conjugates of $U$" +} +{ + "Tag": [ + "\\/closed" + ], + "Problem": "i've been getting some errors today, the only one i thought to copy was this: \r\n\r\nTemplate->loadfile(): File ./templates/PulpFiction/overall_header.tpl for handle overall_header is empty . It was when i click the forum button on the LHS (the frame-like thing).\r\n\r\nanother one was when i posted a message, it said something about e-mails or something, it didnt make much sense.", + "Solution_1": "You must have loaded the page at the same second I uploaded the new template with the Olympiad Math Jam note above. As for other errors, if all goes as planned, we'll be off this garbage server in a week or so." +} +{ + "Tag": [ + "geometry", + "geometric transformation", + "logarithms", + "function", + "rotation", + "IMC", + "college contests" + ], + "Problem": "Let $ f : \\mathbb{R}\\to \\mathbb{R}$ be a continuous function. Suppose that for any $ c > 0$, the graph of $ f$ can be moved to the graph of $ cf$ using only a translation or a rotation. Does this imply that $ f(x) = ax+b$ for some real numbers $ a$ and $ b$?", + "Solution_1": "$ f(x)=e^{x}$ also works: the translation $ (x,y)\\mapsto(x-\\ln c,y)$ maps the grapf of $ f$ onto the graph of $ cf$.", + "Solution_2": "Yes, the problem was pretty easy if you noticed this function. Try doing what I did: assume it's true. Then prove that if such a function exists that is not linear, the corresponding $ G$ must be a horizontal translation. (Unfortunately, I also found an erroneous proof that horizontal translations don't work either....)", + "Solution_3": "hm I was referred to this and this is pretty cool. my idea was to take c \\to 0 and be like well duh if the dude is a translate/rotation of a line, the dude must be a line. the problem is that taking the limit of a path in $ SO(2, \\mathbb{R}) \\ltimes \\mathbb{R}$ need not be defined (did I do the notation for that right?) -- projecting the path into the $ SO(2) \\equal{} S^1$ part yields a limit since the circle is compact, but $ \\mathbb{R}$ is not. However then we realize that the rotation part must be basically trivial (some $ \\varepsilon$ away from some fixed rotation) so you try doing it with just translations and get the exponential" +} +{ + "Tag": [ + "logarithms", + "calculus", + "calculus computations" + ], + "Problem": "find the limit", + "Solution_1": "If you take logarithm:\r\n\\[ \\frac1{\\log x} \\log(1 \\plus{} \\frac 1x) \\sim \\frac {1}{\\log x}\\log{\\frac 1x} \\equal{} \\minus{}1.\\]" +} +{ + "Tag": [ + "calculus", + "integration", + "function", + "calculus computations" + ], + "Problem": "A continuous function $ f(x)$ satisfies that $ \\frac {d}{dx}\\left(\\int_0^x f(x \\plus{} t)\\ dt \\right)\\equal{} 0$. Find the necessary and sufficient condition such that $ \\sum_{n \\equal{} 1}^{\\infty} f(2^n) \\equal{} 1$.", + "Solution_1": "I'm assuming the end parenthesis is supposed to be before the equals sign. At any rate, since f is continuous, it has an antiderivative F'(x) = f(x) and\r\n$ \\frac{d}{dx} (\\int_0^x f(x\\plus{}t)dt) \\equal{} 0 \\Longrightarrow \\frac{d}{dx}(F(x\\plus{}t) \\Big |^{x}_{0}) \\equal{} 0$\r\n$ \\Longrightarrow \\frac{d}{dx}(F(2x) \\minus{} F(x)) \\equal{} 0$\r\n$ \\Longrightarrow 2f(2x) \\minus{} f(x) \\equal{} 0$\r\n$ \\Longrightarrow f(2x) \\equal{} \\frac{1}{2}f(x)$\r\nLetting x = $ 2^n$\r\n$ f(2^{n\\plus{}1}) \\equal{} \\frac{1}{2}f(2^n)$\r\nand the series $ f(2) \\plus{} f(2^2) \\plus{} f(2^3) \\plus{} \\ldots \\equal{} 1$ reduces to\r\n$ f(2) \\plus{} \\frac{1}{2}f(2) \\plus{} \\frac{1}{2^2}f(2) \\plus{} \\ldots \\equal{} 1$\r\n$ f(2) \\plus{} f(2)(\\frac{1}{2} \\plus{} \\frac{1}{2^2} \\plus{} \\frac{1}{2^3} \\plus{} \\ldots) \\equal{} 1$\r\nSince the infinite series in the brackets is equal to 1, the above equation implies that f(2) = $ \\frac{1}{2}$, which is the necessary and sufficient condition such that $ \\sum^{\\infty}_{n\\equal{}1}f(2^{n}) \\equal{} 1$", + "Solution_2": "That's right. You can also answer $ f(1)\\equal{}1$.", + "Solution_3": "Solved it exactly in the same fashion :D" +} +{ + "Tag": [ + "factorial", + "blogs" + ], + "Problem": "Some of you may know this: you pick four positive single-digit integers, and using any operation, exponents, square roots, factorials, etc. you make 24. For instance: 6, 6, 6, 6=6+6+6+6=24. 1, 1, 4, 6=4(6)-1+1=24. If you have a question, feel free to ask.\r\nFirst set:\r\na) 2, 4, 6, 8\r\nb) 3, 3, 3, 3", + "Solution_1": "[quote=\"aopsfan\"]Some of you may know this: you pick four positive single-digit integers, and using any operation, exponents, square roots, factorials, etc. you make 24. For instance: 6, 6, 6, 6=6+6+6+6=24. 1, 1, 4, 6=4(6)-1+1=24. If you have a question, feel free to ask.\nFirst set:\na) 2, 4, 6, 8\nb) 3, 3, 3, 3[/quote]\r\nThis is just for fun. You don't have to sign up.\r\nYou can PM me your answers. You get one point every time you answer correctly. No points if answer is wrong, or of course is not PMd. The first five answers count.", + "Solution_2": "There is an already a 24 marathon currently active. You may wish to see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209924]this thread[/url].", + "Solution_3": "Is there a 100 thread?", + "Solution_4": "no, but if you are thinking on creating it, that would be better into the Math and Strategy Games subforum into the High School forum (where the 24 thread is).", + "Solution_5": "Actually, the standard rules to 24 do not include exponents - only addition, subtraction, multiplication, and division.", + "Solution_6": "[quote=\"pacman2812\"]no, but if you are thinking on creating it, that would be better into the Math and Strategy Games subforum into the High School forum (where the 24 thread is).[/quote]\r\nIm so bad at 100 I don't think there's any use. Just wondering.", + "Solution_7": "I used to play this is in 5th grade! I think it was 7 up or something. Good times good times...\r\n\r\nOnly difference was you had to make 24 using ONLY 7's. :o", + "Solution_8": "Forte that's impossible!", + "Solution_9": "[quote=\"mewto55555\"]Forte that's impossible![/quote]\r\n\r\nI don't remember how it went precicly I think it was making 7. However, the concept was the same. 4 7's and etc. :)", + "Solution_10": "$ (\\frac{7}{7}\\plus{}7)(\\frac{7}{7}\\plus{}\\frac{7}{7}\\plus{}\\frac{7}{7})$", + "Solution_11": "I'm hosting a 24 competition on my blog, the geek's desk, here: [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1219]click here please[/url]. Deadline for enrollment is 19 October 2008. The problems will be tough! Here are the rules: [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=231299]rules[/url]. Basically unlimited competitors. Also, I may have mentioned you can become a contrib for the 24 game--that has changed. You can reply without being a contrib.\r\n\r\nThanks." +} +{ + "Tag": [], + "Problem": "What is the positive square root of the product $ 10 \\times 15 \\times 24$?", + "Solution_1": "Factor to $ 2 \\times 5 \\times 3 \\times 5 \\times 2^3 \\times 3$, so the answer is $ 4 \\times 5 \\times 3\\equal{}\\boxed{60}$.", + "Solution_2": "10=2*5\n24=2*2*2*3\n15=3*5\n10*24*15=2^4*5^2*3^2\nso 10*24*15's square root is 2^(4/2)*5^(2/2)*3^(2/2)=2^2*5^1*3^1=4*5*3=60" +} +{ + "Tag": [ + "function", + "geometry", + "rectangle", + "calculus", + "calculus computations" + ], + "Problem": "A closed rectangular bos with a volume of 16cu. ft is made from 2 kinds of materials. the top and bottom are made of material costing 10cents per square foot and the sides of the material costing 5censts per sqaure foot. what r the dimenstions of the box so that the cost of the materials is minimized.\r\n\r\nHow do I go about starting this.\r\n\r\nI was able to solve a similar one regarding maximizing the volume of a box, nothing with cost though.\r\n\r\nany help is appreciated.", + "Solution_1": "Use Lagrange multipliers.\r\n\r\nLet the dimensions of the box be $x$, $y$, and $z$, where a rectangle with dimensions $x$ and $y$ is the top or bottom.\r\n\r\nNotice that your cost function is $C(x, y, z) = 2(10)xy+2(5)xz+2(5)yz$. The $2$ on each term is the number of sides with that term's area, and the number in parenthesis is the cost per square foot. If you still don't see why this is the cost function, you have a bit of a problem.\r\n\r\nYour constraint curve is $xyz = 16$, which should be obvious from the volume requirement.\r\n\r\nNow just use Lagrange multipliers." +} +{ + "Tag": [ + "calculus", + "integration", + "LaTeX", + "trigonometry", + "function", + "real analysis", + "real analysis unsolved" + ], + "Problem": "Please help with these integrating factor problems.\r\n\r\n1) y' + ln(x)y = 1 y(1)=1\r\nI get stuck here when I have to integrate exp(xlnx - x)\r\n\r\n\r\n2) y' + sin(x)y = sin(x) y(0)=2\r\nI'm working on the integrating factor problem now and although I have the teacher's worked solutions, I'm completely at a loss as to what he's done here...\r\n\r\nfrom: d/dx [ y*exp(-cos(x)) ] = sin(x)*exp(-cos(x))\r\n\r\nto: y(x)*exp(-cos(x)) - y(0)*exp(-cos(0)) = (integral between x,0) of sin(s)*exp(-cos(s)) ds\r\n\r\n\r\ni have no idea how he makes that jump. any suggestions?\r\n\r\n\r\n\r\n3) y' + e*y = x y(1)=2\r\nNot sure what the integrating factor would be in this case...?\r\n\r\n\r\nAlso there are a couple of problems involving the undetermined coefficient method. I'm completely at a loss as to what that is. If someone could explain it I'd be very grateful and could attempt those questions by myself.\r\n\r\n\r\nThanks very much for any help, apologies for not posting in latex format, i tried to make everything clear and legible.", + "Solution_1": "[quote=\"Obstacle1\"]2) y' + sin(x)y = sin(x) y(0)=2[/quote]\r\nLet's find a suitable integrating factor defined by $\\mu (x) = \\exp \\int{\\sin xdx}= e^{-\\cos x}$, this yields\r\n\r\n\\begin{eqnarray*}e^{-\\cos x}y'+\\sin (x)ye^{-\\cos x}&=&e^{-\\cos x}\\sin x\\\\ \\left({e^{-\\cos x}y}\\right)'&=&e^{-\\cos x}\\sin x\\\\ e^{-\\cos x}y&=&\\int{e^{-\\cos x}\\cdot\\sin xdx}\\\\ e^{-\\cos x}y&=&e^{-\\cos x}+c\\\\ y&=&1+ce^{\\cos x}\\end{eqnarray*}\r\n\r\nNow plug the initial condition $y(0)=2$ to find the value of the constant.", + "Solution_2": "[quote=\"Obstacle1\"]\n3) y' + e*y = x y(1)=2\nNot sure what the integrating factor would be in this case...?\n\n[/quote]\r\n\r\n \r\n$y^{'}=\\frac{dy}{dx}$\r\n\r\n$dy+(e^{y}-x)dx=0$\r\n\r\nwell now this is a differential equation in form $M(x,y)dx+N(x,y)dy=0$ i've posted before explanation about this kind of Equations: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133760[/url]\r\n\r\nand now in this problem : $M=1$ and $N=e^{y}-x$\r\n\r\n$M_{y}=0 , N_{x}=-1$\r\n\r\n$N_{x}-M_{x}=-1$\r\nnow let $\\mu(t)$ be the integrating factor and let it be only related to $y$($t=y$), then $\\mu$ will be found by the following formula:\r\n\r\n$\\frac{d\\mu}{\\mu}=\\frac{N_{x}-M_{y}}{M}dy=-dy \\rightarrow \\mu=e^{-y}$\r\n\r\nnow by multiplying $\\mu$ to the equation we'll have:\r\n\r\n$e^{-y}dx-xe^{-y}dy=0$ which is an exact equation Cuz: $M_{y}=N_{x}$\r\n\r\nso there exists a function $u$ such that:\r\n\r\n$du=\\frac{\\partial M}{\\partial x}dx+\\frac{\\partial N}{\\partial y}dy=0$\r\n\r\nnow we have to find $u$,\r\nso: \r\n$u_{x}(x,y)=e^{-y}$\r\n$u_{y}(x,y)=-xe^{-y}$\r\n\r\n$u_{x}(x,y)=e^{-y}\\rightarrow u(x,y)=xe^{-y}+h(y)$\r\n\r\nand so: $u_{y}(x,y)=xe^{-y}+h^{'}(y)$\r\nbut we had: $u_{y}(x,y)=-xe^{-y}$ so $h(y)$ is constant and so:\r\n\r\n$u(x,y)=xe^{-y}$ and because $du=0$ as a consequence the answer is:\r\n\r\n$xe^{-y}=c$\r\n\r\n$y(1)=2$ so $c=e^{-2}$ and the answer is : $xe^{2-y}=1$" +} +{ + "Tag": [ + "function", + "algebra", + "domain", + "calculus", + "integration", + "calculus computations" + ], + "Problem": "Suppose $ f(x) \\equal{} f_{R}(x) \\plus{} f_{I}(x)$ where $ f_{R}(x) \\equal{} 0$ if $ x \\in \\mathbb{Q}$ and $ f_{I}(x) \\equal{}1$ if $ x \\notin \\mathbb{Q}$. Now $ f_R$ nor $ f_I$ are Riemann integrable because they undefined at irrational and rational respectively?", + "Solution_1": "Your topic title says indicator; I suppose you want that $ f$ to be the indicator function on the irrational numbers? That is\r\n\r\n$ f \\equal{} \\begin{cases} 0 \\text{ if } x \\in \\mathbb{Q}\\\\ 1 \\text{ if } x \\not\\in \\mathbb{Q}\\end{cases}$\r\n\r\nTo even begin talking about Riemann integrability you have to specify some domain (say, an interval), on which you would like to integrate (or have some improper integral planned). However, yes, this function is not Riemann integrable, as for any partition $ P \\equal{} \\{x_1, \\dots, x_n\\}$ you will find both rational and irrational numbers in each of the intervals $ [x_i, x_{i\\plus{}1}]$. So the Darboux lower sums are always $ 0$, and the Darboux upper sums are always $ 1$.\r\n\r\n\r\nRegarding about your question about them being \"undefined\". Well, if you have a function $ g : A \\to \\mathbb{R}$ and one $ h : B \\to \\mathbb{R}$ where $ A$ and $ B$ are two disjoint sets (like we have with $ \\mathbb{Q}$ and $ \\mathbb{I}$), then talking about the function $ f$ given by $ f(x) : \\equal{} g(x) \\plus{} h(x)$ does not make sense. This is because, well.... what $ x$'s can be given as an argument to $ f$? Those from $ A$? No, because then $ h(x)$ does not make sense.... same thing for those from $ B$. In short, in order to define a function as the sum of two others, you have to make sure their domains match." +} +{ + "Tag": [ + "blogs", + "\\/closed" + ], + "Problem": "Each time I sign in, I have to go to the URL of my blog before I can use the button. Is there something going on?", + "Solution_1": "Fixed." +} +{ + "Tag": [ + "modular arithmetic", + "graph theory", + "algebra", + "linear equation", + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "Here $G_{n}$ denotes a simple undirected graph with $n$ vertices, $K_{n}$ denotes the complete graph with $n$ vertices, $K_{n,m}$ the complete bipartite graph whose components have $m$ and $n$ vertices, and $C_{n}$ a circuit with $n$ vertices. The number of edges in the graph $G_{n}$ is denoted $e(G_{n})$. \r\n\r\n(a) Let $p$ be a prime. Consider the graph whose vertices are the ordered pairs $(x, y)$ with $x, y \\in\\{0, 1, . . . , p-1\\}$ and whose edges join vertices $(x, y)$ and $(x' , y')$ if and only if $xx'+yy'\\equiv 1 \\pmod{p}$ . Prove that this graph does not contain $C_{4}$ . \r\n(b) Prove that for in\ufb01nitely many values $n$ there is a graph $G_{n}$ with $e(G_{n}) \\geq \\frac{n\\sqrt{n}}{2}-n$ that does not contain $C_{4}$.", + "Solution_1": "Suppose for some $p$ that the given graph does contain a $C_{4}$. Let $(w, w'), (x, x'), (y, y')$ and $(z, z')$ be (in that order) the $C_{4}$. Then manipulating the given relationships, we have \\[w(x-z)+w'(x'-z') \\equiv 0\\equiv y(x-z)+y'(x'-z') \\pmod p.\\] By assumption $(x, x') \\neq (z, z')$, so wlog $x \\neq z$. Then let $-\\frac{x'-z'}{x-z}= k$ and we get that $w \\equiv kw'$ and $y \\equiv ky'$, so $1 \\equiv w'(kx+x') \\equiv y'(kx+x')$ and so $w' \\equiv y'$ and $(w, w') = (y, y')$, contradiction.\r\n\r\nThe given graph has $n = p^{2}$ vertices. A fixed vertex, $(a, b)$, has as neighbors all solutions to the linear equation $ax+by \\equiv 1 \\pmod p$. If $ab \\not\\equiv 0$ then this equation has $p$ solutions (one $y$ for each value of $x$). If $ab \\equiv 0$ but $(a, b) \\neq (0, 0)$ then it has 1 solution. Thus, we have in total $\\frac{1}{2}((p-1)^{2}\\cdot p+2p-2) = \\frac{1}{2}(p^{3}-2p^{2}+3p-2)$ edges. For any prime $p$, this number is larger than $\\frac{p^{3}}{2}-p^{2}= \\frac{n^{3/2}}{2}-n$, so we have an infinite family of graphs satisfying the desired conditions." +} +{ + "Tag": [ + "geometry", + "AMC" + ], + "Problem": "Hey guys, I got some past problems for USAMO/MOP that I posted in the Olympiad section...but i think that it's pretty inactive and plus they're kinda related to the AMC section, so I'll just repost the links and avoid clogging up the latex...cheers!\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126978[/url]\r\n\r\nWarning: the one below has solutions attached, so don't scroll past the problem (i can't edit it for some odd reason...)\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126820[/url]", + "Solution_1": "i responded to both...note the first one, there is not one answer for CE, for the second one...well there is a one-line solution", + "Solution_2": "What do you mean you can't edit it? There should be a button below your post that says \"edit\".", + "Solution_3": "I pushed \"edit\", and it said something along the lines of \"the time to edit this post has passed...\"\r\n\r\nbtw, will adjust problems...", + "Solution_4": "Yeah you can't edit anything more than 24 hours after you post it, unless you tell a mod/admin to do it for you." +} +{ + "Tag": [ + "geometry", + "trigonometry", + "trig identities", + "Law of Cosines" + ], + "Problem": "ABCD is a quadrilateral with mDAB=90 degrees, mBCD=135 degrees, BC=3 and CD= $ 2\\sqrt{2}$. Compute the maximum possible area of ABCD.", + "Solution_1": "[quote=\"firecricket91\"]ABCD is a quadrilateral with mDAB=90 degrees, mBCD=135 degrees, BC=3 and CD= $ 2\\sqrt {2}$. Compute the maximum possible area of ABCD.[/quote]\r\n[hide=\"Solution\"]\nNotice that $ 2[BCD] \\equal{} BC\\cdot CD\\sin \\angle BCD \\equal{} 3\\cdot2\\sqrt {2}\\cdot \\frac {1}{\\sqrt {2}} \\equal{} 6\\implies [BCD] \\equal{} 3$. By the Law of Cosines on $ \\triangle BCD$, we have that\n\\[ BC^2 \\plus{} CD^2 \\minus{} BD^2 \\equal{} 2\\cdot BC\\cdot CD\\cos \\angle BCD\\implies 9 \\plus{} 8 \\minus{} BD^2 \\equal{} 2\\sqrt {2}\\cdot 3\\cdot 2\\cdot \\frac { \\minus{} 1}{\\sqrt {2}}\\implies BD \\equal{} \\sqrt {29}\n\\]\nNow, let $ AB \\equal{} a$ and $ AD \\equal{} b$. The area of $ \\triangle ABD$ is $ \\frac {ab}{2}$. Also, $ a^2 \\plus{} b^2 \\equal{} 29$. Hence, $ 29 \\equal{} a^2 \\plus{} b^2\\ge 2ab \\equal{} 4[ABD]\\implies [ABD]\\le \\frac {29}{4}$. Therefore, $ [ABCD] \\equal{} [ABD] \\plus{} [CDB\\le 3 \\plus{} \\frac {29}{4} \\equal{} \\boxed{\\frac {41}{4}}$. [/hide]" +} +{ + "Tag": [ + "logarithms", + "CEMC", + "AMC", + "USA(J)MO", + "USAMO", + "AoPS Books" + ], + "Problem": "What should one do to prepare for the COMC??", + "Solution_1": "work through comc up until last 2 yrs, try all problems first, then memorize solutions after\r\n\r\nafter prep, take most recent 2 comc's in a mock comc and time yourself and force yourself to write up the written parts", + "Solution_2": "Tips are always helpful unless you're not allowed to do COMC until grade 11, according to NTCI. :?", + "Solution_3": "ask another school", + "Solution_4": "[quote=\"d.lam.86\"]Tips are always helpful unless you're not allowed to do COMC until grade 11, according to NTCI. :?[/quote]\r\n\r\nOk, you go to one messed up school. They won't let you do anything, so find another one. Or at least just write contests at another school", + "Solution_5": "Apparently, COMC is for [i]stronger math students[/i] and I guess their definition of strong math students are the ones in grade 11 and beyond. Nevertheless, I'm not too concerned because I'm definitely not prepared to write it this year. :roll:", + "Solution_6": "write anyways, the first few qs are a joke, you need practice and experience, it will help you later", + "Solution_7": "[quote=\"d.lam.86\"]Apparently, COMC is for [i]stronger math students[/i] [/quote]\n :o \nI thought COMC stands for Canadian [i][b]Open[/b][/i] Mathematics Competition. :huh: \n[quote=\"d.lam.86\"]and I guess their definition of strong math students are the ones in grade 11 and beyond.[/quote]\r\nThis is a screwed, ridiculous definition. :cursing: I can give you tons of counterexample to this. eg) Jacob was grade 9 when he got his first gold medal in IMO. :winner_first:", + "Solution_8": "I get your [i]open[/i] joke but my school doesn't have any exceptional students like the majority of you IMO participants, well at least that's what I think.", + "Solution_9": "[quote=\"d.lam.86\"]I get your [i]open[/i] joke but my school doesn't have any exceptional students like the majority of you IMO participants, well at least that's what I think.[/quote]\r\n\r\nLol, tell your school they could have the potential to have great students if they encouraged it. Besides, just cuz a school doesn't have any exceptional students doesn't mean they never will.", + "Solution_10": "I can understand the reasoning behind their decision-- some of these contests can cost a bundle (well at least my school thinks of it that way) and teachers often refrain from encouraging students to compete in contests that cater to a minorty of math students, and contests they would find moderately expensive, unless of course they have high chances of performing well and bringing positive results to the school, more likely of Grade 11 students than younger ones... This is of course a fraction to the spirit of competition, but you must consider the fact only an exclusive portion of Canadian highschools have entered the math competion pool, many are still catious about the money as well as the importance of this, opposed to other extracurricular activities..", + "Solution_11": "In my first year of high school, I went to my math dept head and asked to sign up for COMC. On the sign up sheet, we were required to write down our grade. When my math head saw \"grade 9,\" he was like \"you know this is for grade 11s and 12s right?\" And I said \"yeah, I've seen the questions.\"\r\n\r\nI scored 7th in my school that year. Although this would be horrible by today's standards, it did impressed my math head quite a bit back then.\r\n\r\nKeep in mind that when I signed up, I didn't even know what logarithms were. Relatively speaking, the month before my first COMC was one of my most \"mathematically-productive\" times. The intense cramming with [i]Problems and How to Solve Them[/i] (published by the CEMC) has really paid off.\r\n\r\nMy advice: go for it. The extra bit of pressure is only going to help you.", + "Solution_12": "[quote=\"billzhao\"]In my first year of high school, I went to my math dept head and asked to sign up for COMC. On the sign up sheet, we were required to write down our grade. When my math head saw \"grade 9,\" he was like \"you know this is for grade 11s and 12s right?\" And I said \"yeah, I've seen the questions.\"\n\nI scored 7th in my school that year. Although this would be horrible by today's standards, it did impressed my math head quite a bit back then.\n\nKeep in mind that when I signed up, I didn't even know what logarithms were. Relatively speaking, the month before my first COMC was one of my most \"mathematically-productive\" times. The intense cramming with [i]Problems and How to Solve Them[/i] (published by the CEMC) has really paid off.\n\nMy advice: go for it. The extra bit of pressure is only going to help you.[/quote]\r\nHi I'm going for the COMC this year too. Yufei, you said something about \"intense cramming\". What do you mean by that? How much did you practice back then? I have went through the books too, but I don't feel like I've improved at all...", + "Solution_13": "Hello. In my school, we pay the $\\$10$ ourselves. Anyway, if you visit [url=http://cemc.uwaterloo.ca/]cemc.uwaterloo.ca/[/url], you will see that it is for grades 10 and above. Well, I'm writing it this year, and I hope spending time in this forum will help me get into CMO or even above CMO. :)\r\n\r\nMasoud Zargar", + "Solution_14": "Wait, Yufei, in 3 years you went from not knowing logs to an imo gold? Thats like impossible.", + "Solution_15": "Where to get this \"Problems and how to solve\" that yufei mentioned?", + "Solution_16": "You can find them at the cemc site https://contest-cemc.uwaterloo.ca/cgi-bin/WebObjects/sigma.woa/wa/orderPubs", + "Solution_17": "I think the Problems and How To Solve Them volumes are popular in Canada; I thought everyone in this Canadian forum knew about them, I guess not [i]everyone[/i] then. I like the CEMC publications because they're challenging for me. At first, I felt they were a bit too hard for me (as in the first volume). They rush really quickly without teaching basics, which is why I've turned to the AoPS books. I like the AoPS books more because they start from the very very basics, which is suitable for a novice like me. I might actually end up doing COMC because I joined the Math club at my school and apparently the whole club is going to write as many competitions as possible, and our head is going to give us a lot of preparation material. I doubt I'll be able to stand 2 1/2 hours for one contest (this will be the first long competition that I'll have *yikes*).\r\n\r\nAre there any specific things COMC markers look for in solutions? I haven't had a lot of experience with writing solutions; let alone any at this point. But the reality is that practice will lead to perfection in my opinion.", + "Solution_18": "lol your school is lucky...our school only allows 25 people to write...", + "Solution_19": "Write the contest. If you have to go to another school, then do it. Just make sure you write the contest. My biggest mistake was not writing this contest. I didn't even know it existed. Honestly, if you can, write COMC and as many other contests as you can. It will get you into camps sooner, which in turn will allow you to improve at math faster. If you wait until grade 11 to start writing contests, you're never gonna be able to make IMO before you graduate.", + "Solution_20": "[quote=\"Elyot\"]Write the contest. If you have to go to another school, then do it. Just make sure you write the contest. My biggest mistake was not writing this contest. I didn't even know it existed. Honestly, if you can, write COMC and as many other contests as you can. It will get you into camps sooner, which in turn will allow you to improve at math faster. If you wait until grade 11 to start writing contests, you're never gonna be able to make IMO before you graduate.[/quote]\r\n\r\nBut but ..but ...Elyot, didn't you make IMO after basically starting to write contests in Grade 11? :roll:", + "Solution_21": "he's special", + "Solution_22": "[quote=\"TheDreamer\"][quote=\"Elyot\"]Write the contest. If you have to go to another school, then do it. Just make sure you write the contest. My biggest mistake was not writing this contest. I didn't even know it existed. Honestly, if you can, write COMC and as many other contests as you can. It will get you into camps sooner, which in turn will allow you to improve at math faster. If you wait until grade 11 to start writing contests, you're never gonna be able to make IMO before you graduate.[/quote]\n\nBut but ..but ...Elyot, didn't you make IMO after basically starting to write contests in Grade 11? :roll:[/quote]\r\n\r\nNo no no... I wrote Euclid every year since grade 8... Descartes in grade 9... I had to pester my school like mad to let me write the open... I made group 2 in CMO, then the next year I got to go to winter camp, write APMO and USAMO, and they picked me for IMO from that. Seriously, grade 11 is like... your LAST CHANCE. I was lucky that I made the team, because there was some insanely tough competition for spots last year. This year will be even worse (which is unfortunate for those of you who are graduating this year) but the year after this will be much better, as Yufei, David, Richard, and Peng will be gone.\r\n\r\nBut yeah... the earlier you can write CMO, the better.\r\n\r\nI still wish I had a year more of practice... I could have done a lot better than I did.", + "Solution_23": "I signed up for COMC yesterday(Is my school late? :? ).\r\n\r\nWrite COMC!! This is only way to let the people, who select the team, know that you exist! :)", + "Solution_24": "Skip that thought...", + "Solution_25": "The deadline is November 9th. Thank god, I thought I was too late." +} +{ + "Tag": [ + "analytic geometry" + ], + "Problem": "A segment with endpoints at $ A(2, \\minus{}2)$ and $ B(14, 4)$ is extended\nthrough $ B$ to point $ C$. If $ BC \\equal{} (\\frac{1}{3})AB$, what are the coordinates for point $ C$? Express your answer as an ordered pair.", + "Solution_1": "The $ x$-coordinate increases by $ \\frac{14\\minus{}2}{3}\\equal{}4$, and the $ y$-coordinate increases by $ \\frac{4\\minus{}(\\minus{}2)}{3}\\equal{}2$. Thus, the coordinates for point $ C$ are $ (14\\plus{}4,4\\plus{}2)$ or $ \\boxed{(18,6)}$." +} +{ + "Tag": [ + "AMC", + "AIME", + "USA(J)MO", + "USAMO", + "geometry", + "ARML" + ], + "Problem": "I am a high school teacher and three of my students qualified for the AIME; their scores were 3,4 and 5.\r\nThe ironic thing is that the hypotenuse was a Freshman and the longest leg a junior.\r\n\r\nTwo of my students put a bit of time into preparing for the AIME on their own (and I pushed them gently to prep).\r\nThe students that have reached the USAMO level know just how much knowledge and experience it takes (and work!). \r\n\r\nThere is a HUGE amount of print and on-line resources for students to use to prepare on their own, but here is my question:\r\nHow many students that reach the top level get a chance to prepare with other such students IN PERSON?\r\nWhat types of in person groups have been formed? There are a few exceptional schools that have a number of USAMO qualifiers, but for many of these students, they are the only individuals in their schools to make it.\r\n\r\nI am from Nassau County, NY, and I'm wondering what structure exists in other counties.\r\n\r\nNassau had two USAMO qualifiers this year and quite a number of students that could potentially qualify next year if they set their minds to it. I'm wondering if we could organize an in person problem solving group for students to teach each other techniques and work on problems together. Maybe it could meet once a week starting in the Summer or Fall; I could help get it started, but perhaps the group could really be run by the students. This could definetly work if the kids had the right resources.\r\n\r\nIs any one from Nassau County reading this - are you iterested?\r\n\r\nHave other people done similar things in other counties?", + "Solution_1": "You're right. It can be so much harder if you're the \"only one.\" The internet is certainly one possibility, especially because it offers so much flexibility. However, another extremely important component to preparing for the AIME is to encourage students to struggle with problems on their own for long periods of time. Just throwing out random thoughts.", + "Solution_2": "Yeah, I've always been the \"only one.\" It seems that the good math students in my state are too (TN only had three qualifiers, one on the other side of the state.) Now that i've learned about the AIME and USAMO, I was actually planning to get more students interested in the area. Hopefully we can get some kind of math club/league like the NYSML and others. I think if you want to have other students similarly motivated, there has to be someone to get the others into it. There's one aops-er who qualified for mosp as a freshman last yr, made usamo this yr, besides a perfect score on the amc 10b. He runs and coaches the math club at his school, but still he's the only one to advance beyond the AIME (don't mean to embarass you sam). I just think if you want more ppl into competitive math, you might just have to drag them into it :roll:", + "Solution_3": "Things are really different where I live. I personally know 9 kids who qualified for USAMO this year, and these are all kids in the local area who I have taken summer classes with.", + "Solution_4": "Have them join the regional ARML team." +} +{ + "Tag": [ + "analytic geometry", + "geometry", + "3D geometry", + "sphere", + "vector" + ], + "Problem": "a skier skis from the top of a snowball down the side, in a uniform gravitational field. at what point on the surface of the snowball, given by the angle of the radius at that point to the vertical, does the skier lose contact with the surface of the snowball?", + "Solution_1": "I believe we had a similar problem in our school leaving examinations in physics last year. It shouldn't be too difficult if the solution is what I think :)...", + "Solution_2": "ok cheers... just kinda looking for hints:\r\n\r\ndid you use energy or momentum conservation to calculate it?\r\ndid you need the formula for acceleration in polar coordinates?", + "Solution_3": "You only need a few things.\r\n1) Understand that the normal force is $0$ when you lose contact\r\n2) Centripetal acceleration is $v^2/r$\r\n3) Conservation of energy (that's how you get $v$)\r\n4) It would be easiest to solve in terms of the angle when you lose contact force", + "Solution_4": "Yeah, kind of what FMako says....", + "Solution_5": "but can you use v^2/r.. the angular velocity is increasing so I thought you'd need to use a more general form of circular motion..", + "Solution_6": "[quote=\"degreestudent\"]but can you use v^2/r.. the angular velocity is increasing so I thought you'd need to use a more general form of circular motion..[/quote]\r\n\r\nUse $\\frac{v^2}{r}$ just for the moment where skier loss contact, i.e. when N=0.\r\nDoing this you will get equations which you can solve.", + "Solution_7": "yes, the angular velocity is increasing, but while the skier is touching the hemisphere, well, the velocity vector's tangent to the surface, yes?\r\n\r\nSo draw yourself a free body diagram...", + "Solution_8": "At the point of breaking off the sphere N=0;\r\n\r\nOn solving you will find that the body breaks off from a point at which the radius vector makes an angle arccos(2/3) with the vertical.", + "Solution_9": "IIRC this was an Irodov problem, slightly differently put of course." +} +{ + "Tag": [ + "Putnam", + "algebra", + "polynomial", + "college contests" + ], + "Problem": "Let $ n_10$, then\r\n\r\n$ \\frac{ab\\plus{}3c}{a\\plus{}b\\plus{}2c}\\plus{}\\frac{ca\\plus{}3b}{a\\plus{}2b\\plus{}c}\\plus{}\\frac{bc\\plus{}3a}{2a\\plus{}b\\plus{}c}\\leq\\frac{a\\plus{}b\\plus{}c\\plus{}9}{4}$", + "Solution_1": "[quote=\"filomen\"]If $ a,b,c > 0$, then\n\n$ \\frac {ab \\plus{} 3c}{a \\plus{} b \\plus{} 2c} \\plus{} \\frac {ca \\plus{} 3b}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {bc \\plus{} 3a}{2a \\plus{} b \\plus{} c}\\leq\\frac {a \\plus{} b \\plus{} c \\plus{} 9}{4}$[/quote]\r\n\r\nlet $ a\\plus{}b\\plus{}c\\equal{}3u, ab\\plus{}bc\\plus{}ca\\equal{}3v^2, abc\\equal{}w^3$, then ineq becomes\r\n\r\n$ f(w^3)\\equal{}(63u\\minus{}27)w^3\\plus{}162u^4\\minus{}189u^2v^2\\minus{}36v^4\\plus{}162u^3\\minus{}135uv^2 \\ge 0$\r\n\r\nfix $ u,v$, that we just need to prove when $ w^3$ get max and min the ineq holds, i.e. just need to prove when $ b\\equal{}c$ and $ a \\equal{} 0$ the ineq holds.\r\n\r\n(1) when $ b\\equal{}c$, ineq can writen as $ \\frac{(a\\plus{}3)(a\\minus{}c)^2}{4(a\\plus{}3c)(a\\plus{}c)} \\ge 0$, clear true;\r\n\r\n(2) when $ a\\equal{}0$, ineq can writen as $ \\frac{3(b\\plus{}c)(2b^2\\minus{}bc\\plus{}2c^2)\\plus{}(b\\plus{}2c)(c\\plus{}2b)(b\\minus{}c)^2}{4(b\\plus{}2c)(c\\plus{}2b)(b\\plus{}c)}\\ge0$, clear true.\r\n\r\ndone.", + "Solution_2": "[quote=\"filomen\"]If $ a,b,c > 0$, then\n\n$ \\frac {ab \\plus{} 3c}{a \\plus{} b \\plus{} 2c} \\plus{} \\frac {ca \\plus{} 3b}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {bc \\plus{} 3a}{2a \\plus{} b \\plus{} c}\\leq\\frac {a \\plus{} b \\plus{} c \\plus{} 9}{4}$[/quote]\r\nIt's obviously trues by CS and notice that \r\n$ \\sum \\frac {ab}{a \\plus{} b \\plus{} 2c} \\le \\frac {a \\plus{} b \\plus{} c}{4}$ ; $ \\sum \\frac {c}{a \\plus{} b \\plus{} 2c} \\le \\frac {3}{4}$ :)" +} +{ + "Tag": [ + "combinatorics unsolved", + "combinatorics" + ], + "Problem": "Let $ P$ be a polyhedron with $ s$ faces. Prove that it contains at least $ s/7 \\plus{} 1$ faces with equal number of sides.", + "Solution_1": "and what is $ l$?", + "Solution_2": "Sorry, $ s\\equal{}l$, the number of faces. I'll correct it.", + "Solution_3": "let $ f_k$ denote the number of faces having $ k$ sides, and let $ V, E, F$ denote the number of vertices, edges, and faces, respectively. let $ n$ be the largest number of sides that any face has. we have\r\n\r\n(1) $ \\sum_{k=3}^n f_k= F$\r\n(2) $ \\sum_{k=3}^n \\frac{k}{2}f_k= E$. (each edge is counted in exactly 2 faces.) \r\n(3) $ \\sum_{k=3}^n \\frac{k}{3} f_k \\geq V$. (each vertex is counted in at least 3 faces.)\r\n\r\nby Euler's Theorem, $ V-E+F=2$, so $ \\sum_{k=1}^n \\left(1-\\frac{k}{6}\\right)f_k \\geq 2$. the result is clear for $ n\\leq 8$. now, assume that $ n\\geq 9$ and that $ f_k \\leq \\frac{F}{7}$ for all $ k$. then\r\n\r\n\\[ 2 \\leq \\sum_{k=3}^n \\left(1-\\frac{k}{6}\\right) \\frac{F}{7}\\\\\r\n\\&= \\frac{F}{14}+\\sum_{k=9}^n\\left(1-\\frac{k}{6}\\right) \\frac{F}{7}\\\\\r\n\\&\\leq \\frac{F}{14}-(n-8)\\frac{F}{14}\\\\\r\n\\&=\\frac{9F-Fn}{14}\\]\r\n\r\nso $ 28\\leq 9F-Fn$, i.e. $ n\\leq \\frac{9F-28}{F}<\\frac{9F}{F}=9$, a contradiction" +} +{ + "Tag": [ + "logarithms", + "integration", + "calculus", + "calculus computations" + ], + "Problem": "hello folks :D \r\n\r\nshow\r\n\r\n$ \\beta'(1) \\;\\equiv\\; \\sum_{k\\equal{}1}^\\infty\\;\\dfrac{(\\minus{}1)^{k\\plus{}1}\\cdot\\ln\\,(2k\\plus{}1)}{2k\\plus{}1}$\r\n\r\n$ \\;\\equal{}\\; \\boxed{\\beta(1)\\cdot\\left(3\\ln\\pi\\plus{}\\ln\\,4\\plus{}\\gamma\\minus{}4\\ln\\left(\\Gamma\\left(\\dfrac{1}{4}\\right)\\right)\\right)}$", + "Solution_1": "Differentiate the formula $ \\beta(\\alpha)\\, \\Gamma(\\alpha)\\equal{}\\frac12 \\int_0^\\infty \\frac{x^{\\alpha\\minus{}1}}{\\cosh x}\\, dx$ to obtain\r\n\r\n$ \\beta'(\\alpha)\\, \\Gamma(\\alpha)\\plus{}\\beta(\\alpha)\\, \\Gamma'(\\alpha)\\equal{}\\frac12 \\int_0^\\infty \\frac{x^{\\alpha\\minus{}1}\\, \\ln x}{\\cosh x} \\, dx$ and set $ \\alpha\\equal{}1$ .\r\n\r\n$ \\beta'(1)\\minus{}\\beta(1)\\, \\gamma\\equal{}\\frac12 \\int_0^\\infty \\frac{\\ln x}{\\cosh x} \\, dx\\equal{}\\frac{\\pi}{2}\\, \\ln\\left(\\sqrt{2\\pi} \\, \\frac{\\Gamma\\left(\\frac34\\right)}{\\Gamma\\left(\\frac14\\right)}\\right)$ ([url=http://www.mathlinks.ro/viewtopic.php?t=164417]Vardi's integral[/url])\r\n\r\nAfter Legendre's duplication formula $ \\Gamma\\left(\\frac34\\right)\\equal{}\\Gamma\\left(\\frac14\\plus{}\\frac12\\right)\\equal{}\\sqrt{2\\pi}\\, \\frac{\\Gamma\\left(\\frac12\\right)}{\\Gamma\\left(\\frac14\\right)}$\r\n\r\nHence Vardi's integral can be written as $ \\frac{\\pi}{2} \\ln\\left(2\\pi \\,\\frac{\\sqrt{\\pi}}{\\Gamma^2\\left(\\frac14\\right)}\\right)\\equal{} \\frac{\\pi}{4} \\ln\\left(\\frac{4\\pi^3}{\\Gamma^4\\left(\\frac14\\right)}\\right)$\r\n\r\n$ \\equal{}\\beta(1)\\, \\left(\\ln 4\\plus{}3\\ln \\pi\\minus{}4\\ln\\Gamma\\left(\\frac14\\right)\\right)$ ." +} +{ + "Tag": [ + "inequalities", + "function", + "inequalities unsolved" + ], + "Problem": "Let $x,y,z>0$ and $xyz=1$.Prove that:\r\n$\\sqrt{\\frac{x}{1+xz}}+\\sqrt{\\frac{y}{1+yx}}+\\sqrt{\\frac{z}{1+zy}}\\ge \\sqrt{\\frac{16(1+x+xz)(1+y+yx)(1+z+zy)}{3(1+xz)(1+yx)(1+zy)}}$.", + "Solution_1": "If i'm not wrong, you can homogenize all, and use jensen for the convex function $\\frac{1}{\\sqrt{x}}$!\r\n\r\n(I think that this is a different form of one of the problems in the mathematical reflexions, issue 2 2007)" +} +{ + "Tag": [ + "inequalities" + ], + "Problem": "Let x, y, and z belong to the real numbers.\r\n\r\nIf x > 3, y < 5, and z < 4, then prove:\r\n\r\n$ x^2 \\minus{} y \\minus{} z > 0$.\r\n\r\n\r\n*** EDIT: \r\n\r\nAlternate solution to KrazyFK's:\r\n[hide]\nLet $ \\epsilon_1, \\epsilon_2, \\epsilon_3$ belong to the real numbers, and\nlet $ \\epsilon_1, \\epsilon_2, \\epsilon_3 > 0$.\n\nLet $ x \\equal{} 3 \\plus{} \\epsilon_1$\nLet $ y \\equal{} 5 \\minus{} \\epsilon_2$\nLet $ z \\equal{} 4 \\minus{} \\epsilon_3$\n\n$ x^2 \\minus{} y \\minus{} z$ becomes:\n\n$ (3 \\plus{} \\epsilon_1)^2 \\minus{} (5 \\minus{} \\epsilon_2) \\minus{} (4 \\minus{} \\epsilon_3)$\n\n$ 9 \\plus{} 6\\epsilon_1 \\plus{} (\\epsilon_1)^2 \\minus{} 5 \\plus{} \\epsilon_2 \\minus{} 4 \\plus{} \\epsilon_3$\n\n$ 6\\epsilon_1 \\plus{} (\\epsilon_1)^2 \\plus{} \\epsilon_2 \\plus{} \\epsilon_3 > 0$.\n\nBecause the left side is always positive, this inequality is true.\n\nTherefore, the original inequality is true which was to be proven.\n\n[/hide]", + "Solution_1": "Umm... this seems too easy.\r\n\r\nIf $ x > 3$, then $ x^2 > 9$\r\n\r\n$ x^2 \\minus{} y \\minus{} z \\equal{} x^2 \\minus{} (y \\plus{} z)$\r\n\r\nHowever, since $ y < 5$ and $ z < 4$ then $ (y \\plus{} z) < 9 < x^2$\r\n\r\nIt immediately follows that $ x^2 \\minus{} y \\minus{} z > 0$" +} +{ + "Tag": [ + "induction", + "modular arithmetic" + ], + "Problem": "how about this one ? I have been trying for hours...but still couldn't get it...\r\n\r\n\r\nProve that 21 divides 4^(n+1) + 5^(2n-1)\r\n\r\nhow do you prove this one by induction ???[/img]", + "Solution_1": "Direct method:\r\n[hide]\nSubstitute $ n$ with $ m \\plus{} 1$, where $ m$ is nonnegative integers (I assume n is positive integers), and then\n$ 4^{n \\plus{} 1} \\plus{} 5^{2n \\minus{} 1}$\n$ \\equal{} 4^{m \\plus{} 2} \\plus{} 5^{2m \\plus{} 1}$\n$ \\equal{} 16\\cdot 4^m \\plus{} 5\\cdot 25^m$\n\nAnd since $ 16\\equiv \\minus{} 5\\pmod{21}$, and $ 25\\equiv4\\pmod{21}$, there follows,\n$ 16\\cdot 4^m \\plus{} 5\\cdot 25^m\\equiv \\minus{} 5\\cdot 4^m \\plus{} 5\\cdot 4^m\\equiv0\\pmod{21}$\nThus, the original expression is always divisble by $ 21$.\n[/hide]", + "Solution_2": "[hide=\"induction?\"]For 1, it is true, since 21|21. Assume it holds for $ n$. For $ n\\plus{}1$, \n$ 4^{n\\plus{}2}\\plus{}5^{2n\\plus{}1}$ is\n$ 16(4^n)\\plus{}5(25^n)$/\nIn modulo 25, this is $ \\minus{}5*4^n\\plus{}5*4^n\\equiv0\\pmod{21}$, and we're done.[/hide]\r\nEssentially the same as the direct case, so I think I messed up.", + "Solution_3": "[hide=\"This is the inductive solution\"] Base case: $ n \\equal{} 1$. Then $ 4^2 \\plus{} 5 \\equal{} 21$. \n\nInductive step: If $ 21 | 4^{k\\plus{}1} \\plus{} 5^{2k\\minus{}1}$ then\n\n$ 4^{k\\plus{}2} \\plus{} 5^{2k\\plus{}1} \\equal{} 4 \\cdot 4^{k\\plus{}1} \\plus{} 25 \\cdot 5^{2k\\minus{}1} \\equal{} 4(4^{k\\plus{}1} \\plus{} 5^{2k\\minus{}1}) \\plus{} 21(5^{2k\\minus{}1})$\n\nWe know that $ 21$ divides both terms (by our inductive hypothesis), so $ 21$ divides $ 4^{k\\plus{}2} \\plus{} 5^{2k\\plus{}1}$. [/hide]" +} +{ + "Tag": [ + "function", + "real analysis", + "real analysis unsolved" + ], + "Problem": "$f$ is a real fuction and bounded in $[a,b]$, $\\alpha$ is non-decreasing function in $[a,b]$. Defind :\r\nSet $P=\\{x_0,x_1,...,x_n\\}$, $x_0=a,x_n=b$ and $|P|=\\max_{i=1,n}(x_i-x_{i-1})$\r\n$L(P,f,\\alpha)=\\sum_{i=1,n}m_i(\\alpha(x_i)-\\alpha(x_{i-1}))$ , $m_i=\\inf(f(x)|x\\in [x_{i-1},x_i]$\r\n$r=\\sup L(P,f,\\alpha)$ for all $P$ in $[a,b]$ .\r\n\r\nProve that for all $\\epsilon >0$, exists $\\gamma >0$ st :\r\n\r\n$r-L(P,f,\\alpha) \\le \\epsilon$ for all $P$ st $|P| \\le \\gamma$.", + "Solution_1": "Every monotone function has a numerable set of discontinities, and being bounded is integrable. Follows your conclusion.", + "Solution_2": "[quote=\"xirti\"]Every monotone function has a numerable set of discontinities[/quote]\r\n\r\nHow do you prove it?", + "Solution_3": "Every monotone ( I assume increasing) function has lateral limits in every point. Define for every $x$ $g(x)=f(x-)$ and $h(x)=f(x+)$. Since $f$ is monotone, intervals of the form $(g(x);h(x))$ are disjoint. So, in $R$ there is a nenumerable reunion of disjoint intervals. But we know that in $R$ every set of disjoint intervals is numerable:\r\n\r\n Since in every interval lies a rational number we can define o injection from that set to $\\mathbb{Q}$.", + "Solution_4": "Just to be sure I understood your proof: You used the \"reductio ad absurdum\" method, right?", + "Solution_5": "Yes. A stronger result holds: For any function the set of jump discontinuities is numerable. But I think that it hasn't an elementary proof.", + "Solution_6": "Notice that $f(x)$ is a abiraty real function in [a,b], non condition increasing. Only $\\alpha(x)$ is non-decreasing.", + "Solution_7": "Sorry :D I didn't read the text well because it was too long. \r\n\r\nI have a solution: \r\n\r\nConsider $P$ a fixed division of interval $[a,b]$ such that $r-L(P,f,\\alpha)<\\frac \\epsilon 2$. Consider another division $Q$ such that $|Q|<\\niu 1$ is prime if and only if it is not expressible as a sum of three or more consecutive positive integer.", + "Solution_1": "I m surprised that no one solved it :huh:", + "Solution_2": "well... this is what i get\r\n\r\nas the same thing the other person did, if $i$ can be express as the sum of consecutive integer, then \r\n\r\n$\\displaystyle i = \\frac{(2n + m)(m + 1)}{2}$\r\n\r\nso\r\n\r\n$2i = (2n + m)(m + 1)$\r\n\r\nwhere one of $2n + m$ or $m + 1$ must be even.\r\n\r\nlets suppose that $i$ is any prime number except $2$, then one of the factor must be $1$\r\nbut because $m \\ge 2$ since $i$ is a sum of $3$ or more consecutive integers, so $2n + m \\ge 4$ and $m + 1 \\ge 3$\r\n\r\nas we can see, $2n + m$ and $m + 1$ will always be greater than $2$, so it is impossible to have a prime with a sum of 3 or more consecutive integers.\r\n\r\nso now we have to prove that an odd number can always be express as the sum of three or more consecutive integers. which means, $2i = (2n + m)(m + 1)$ holds true for $m \\ge 2$ and odd $i$, which is not a prime\r\n\r\nwe can factor $i$ as $i = pq$, with $p$ and $q$ as odd numbers other than $1$, so $2i = 2p * q$, so now $p$ is even, $q$ is odd. now our proof becomes that one can find $2p$ and $q$ such that $2p|2i$ and $q|2i$ for any $i$\r\n\r\n\r\nwe can choose $m$ so that $m+1 = 2p$ or $m+1 = q$ , we can be sure that the $m$ choosen will definitely be greater than or equal to $2$ because $p,q \\ge 3$\r\n\r\nnow, all we have to do is prove that there must be a factor of $2i$, $2n-1$ units away, for any positive integer $n$, and $(2n + m)(m + 1) = 2i$\r\n\r\nwe found that $2n-1$ for positive integer $n$ yields all odd integer. so if $m + 1$ is odd, $2n + m$ will be even. and if $m + 1$ is even, $2n + m$ is odd. so $2n + m$ can be all possible odd or even integer, for which one must satisfy $(2n + m)(m + 1) = 2i$, therefore the problem is proved.\r\n\r\np.s: m+1 must be, and always can be, chosen so that it is the lesser factor" +} +{ + "Tag": [ + "geometry", + "symmetry", + "integration", + "trigonometry", + "calculus", + "calculus computations" + ], + "Problem": "Find the area of the region that lies inside the cardioid r = 1 + cos(theta) and outside the circle r = 1.", + "Solution_1": "Definitely graph this. You will see that there is some symmetry. The two curves meet at $ \\theta \\equal{} \\pm\\frac{\\pi}{2}$. Due to symmetry we can just find the area in the first quadrant and multiply by $ 2$ though.\r\n\r\n$ \\frac{1}{2}A\\equal{}\\frac{1}{2}\\int_0^\\frac{\\pi}{2} r^2d\\theta \\minus{}\\frac{\\pi}{4}$ so\r\n\r\n$ A\\equal{}\\int_0^\\frac{\\pi}{2} (1\\plus{}\\cos\\theta)^2d\\theta\\minus{}\\frac{\\pi}{2}\\equal{} \\int_0^\\frac{\\pi}{2}(1\\plus{}2\\cos\\theta\\plus{}\\cos^2\\theta)d\\theta\\minus{}\\frac{\\pi}{2}$\r\n\r\nThe rest is integration. If you need help on the rest just say so. Oh yeah, so I subtracted the area inside the circle in the first quadrant without the integral ($ \\frac{\\pi}{4}$)." +} +{ + "Tag": [ + "function", + "Olimpiada de matematicas" + ], + "Problem": "[color=blue]Intent\u00e9 resolver un problema de ecuaciones funcionales. Pude demostrar que una funci\u00f3n se cumple para los racionales, y verifica si se cumple con los reales, pero no logr\u00e9 hallar una buena demostraci\u00f3n para ese caso.\n\u00bfAlguien podr\u00eda ayudarme, proporcionando los distintos m\u00e9todos para lograrlo?[/color]", + "Solution_1": "Seria de ayuda q postees el enunciado del problema!!\r\n\r\n\r\nSaludos,", + "Solution_2": "[color=blue]Sea $f: \\mathbb{R}\\to \\mathbb{R}$, que cumple la ecuaci\u00f3n $\\left[{f\\left( x \\right)+f\\left( z \\right)}\\right]\\left[{f\\left( y \\right)+f\\left( t \\right)}\\right] = f\\left({xy-zt}\\right)+f\\left({xt+yz}\\right)$.\n\nEl enunciado pide hallar todas las funciones que cumplan la ecuaci\u00f3n. Dos de ellas son $f\\left( x \\right) = 0$ y $f\\left( x \\right) = \\tfrac{1}{2}$. Descartando ambos casos, encontr\u00e9 que $f\\left( x \\right) = x^{2}$ tambi\u00e9n verifica, pero s\u00f3lo demostr\u00e9 para todo $x \\in \\mathbb{Q}$. \u00bfC\u00f3mo logro demostrar para todo $x \\in \\mathbb{R}$[/color]", + "Solution_3": "[color=blue]O\u00ed hablar de un m\u00e9todo de $Cauchy$ que podr\u00eda ayudarme a solucionar lo que me queda del enunciado.\n\u00bfQui\u00e9n podr\u00eda explicarme adecuadamente c\u00f3mo se trabaja con este m\u00e9todo?[/color]", + "Solution_4": "Vease la resolcion de ese problema en la revista eureka #15 \r\nhttp://www.obm.org.br/frameset-eureka.htm\r\n\r\nEl criterio que se utiliza es muy bueno dado que si la funcion satisface para los racionales y ademas es creciente entonces satisface para todos los reales . \r\n\r\nSaludos", + "Solution_5": "si, es bien sabido q si la funcion satisface un valor para los racionales y ademas es continua, entonces la funcion tamb cumple q toma ese valor para los reales... en especial la monotonia (creo q ese es el termino) garantiza q la funcion sea continua.", + "Solution_6": "Aprovecho este topic para postear el siguiente link: http://www.imocompendium.com/tekstkut/funeqn_mr.pdf", + "Solution_7": "[quote=\"campos\"]en especial la monotonia (creo q ese es el termino) garantiza q la funcion sea continua.[/quote]\r\n\r\nOjo... Mucho ojo con hacer afirmaciones tan a la ligera.\r\n\r\nConsideremos por ejemplo la funci\u00f3n $ f(x)$ definida como sigue:\r\n\r\n$ f(x) \\equal{} x$ si $ x < 0$\r\n$ f(0) \\equal{}\\frac{1}{2}$\r\n$ f(x) \\equal{} x\\plus{}\\frac{1}{2}$ si $ x > 0$\r\n\r\nEsa funci\u00f3n es claramente mon\u00f3tona pero no es continua. Se sabe de hecho que el conjunto de discontinuidades de una funci\u00f3n mon\u00f3tona puede ser a lo m\u00e1s numerable." +} +{ + "Tag": [ + "counting", + "distinguishability", + "floor function" + ], + "Problem": "[b]Find the number of non-negative quadruples of integers (a,b,c,d) that satisfy a+b+c+d=3.\nFind the number of non-negative quadruples of integers (a,b,c,d) that satisfy a+b+c+d=4[/b]\r\n\r\nIntuitively, the solutions to each problem should be $ \\binom{6}{3}$ and $ \\binom{7}{3}$ respectively. However, if you make a one-to-one correspondence where for the 1st problem you are trying to find the number of ways to place 3 indistinguishable marbles in 4 distinguishable boxes, you get 10, not the 20 you would get using the balls-and-urns method. Is there something wrong with that correspondence?", + "Solution_1": "How do you count the number of ways to place 3 indistinguishable marbles in 4 distinguishable boxes without listing them out as ordered quadruples? \r\n\r\nAlso, this isn't answering your question directly, but could you explain the balls and urn method for this particular problem to me? How is it like the \"make the problem into a counting letters problem method\" illustrated below?\r\n\r\nFor the first problem, let's say that we have three circles to correspond to the three on the RHS. We want to divide it into four groups. Let's represent dividers of the groups as \"D's\" and circles as \"C's\"\r\n\r\nSo one possible partition (wrong word?) of three would be \r\nDDCCDC. We realize that counting the orders of the letters is the same as counting the solutions so we have six choose three solutions. \r\n\r\nLastly, is it easier to think of problems as balls and urns problems or as arrangements of some distinguishable and indistinguishable letters?", + "Solution_2": "[quote=\"mihail911\"]the number of ways to place 3 indistinguishable marbles in 4 distinguishable boxes, you get 10,[/quote]\r\n\r\nCare to show your work here?\r\n\r\nEdit: Perhaps you accidentally calculated $ {5 \\choose 2}$? Given three marbles and three dividers, there are clearly $ {6\\choose 3} \\equal{} 20$ ways.", + "Solution_3": "How would you think of the problem in a balls/urns way?", + "Solution_4": "Wait, he didn't say how he got $ 10$. I guessed that he accidentally calculated $ {6 \\minus{} 1 \\choose 3 \\minus{} 1} \\equal{} {5\\choose 2}$. Edit: Ignore this sentence. \r\n\r\nNote that in a quadruple, order matters. So consider four urns, $ \\lfloor a \\rfloor\\ \\lfloor b \\rfloor\\ \\lfloor c \\rfloor\\ \\lfloor d \\rfloor$; we want to drop three \"balls\" into these urns. Can you confirm via a bijection that we can draw this correspondence to the ball and urn argument?\r\n\r\n(For example, $ \\lfloor\\ \\rfloor\\ \\lfloor \\circ \\circ \\rfloor\\ \\lfloor\\ \\rfloor\\ \\lfloor\\ \\circ \\rfloor$ and $ \\lfloor\\circ\\circ\\circ \\rfloor\\ \\lfloor \\ \\rfloor\\ \\lfloor\\ \\rfloor\\ \\lfloor\\ \\rfloor$ respectively correspond to the quadruples $ (0,2,0,1), \\ (3, 0, 0, 0)$. We can tell pretty easily that we can represent any such $ a,b,c,d$ in this manner).\r\n\r\nSo we are dropping three balls into four containers. To count the number of ways this is possible, consider making the containers into just the dividers between the containers. We have a string of $ |\\circ\\circ|\\ |\\circ$ or something along those lines then; we see that there must be $ 3\\ \\circ$s and $ 3\\ |$s, so there are $ {6\\choose 3}$ ways to do this.\r\n\r\nThe correspondence is perfectly fine. mihail911, you probably used the wrong formula.\r\n\r\nAs to how you would think of this as a ball-and-urn problem, mostly because the method behind the problem $ a\\plus{}b\\plus{}c \\plus{} \\cdots \\equal{} k$ is well-known." +} +{ + "Tag": [ + "algebra", + "polynomial", + "linear algebra", + "matrix", + "linear algebra theorems" + ], + "Problem": "Let $A\\in\\mathcal M_2(R)$ and $P\\in R[X]$ with no real roots. If $\\det(P(A))=0$ then $P(A)=O_2$.\r\n\r\n Can this problem be generalised?", + "Solution_1": "Nope .Example:for $A \\in M_n(R)$ that has its minimal $\\pi=(X^2+X+1)*(X-1)$ consider $P=X^2+X+1$", + "Solution_2": "You can't do that- we're restricting to $2\\times 2$ matrices.\r\n\r\nThe generalized version:\r\nLet $F$ be a field, and $p$ be a polynomial over $F$ such that all irreducible factors of $p$ have degree at least $n$. Then if $M$ is an $n\\times n$ matrix over $F$ and $p(M)$ is singular, $p(M)=0$.\r\n\r\nWe have to change fields in the generalization because there are no irreducible polynomials with degree greater than 2 over $\\mathbb{R}$.", + "Solution_3": "For the original problem:\r\n\r\nLet $\\lambda_i, \\overline{\\lambda_i} \\in \\mathbb C \\setminus \\mathbb R$ be the roots of $P$.\r\n\r\nThen $\\det P(A) = \\prod \\det \\left( A - \\lambda_i I_2 \\right) \\det \\left( A - \\overline{\\lambda_i} I_2 \\right) = 0$.\r\n\r\nThus, $A$ has a non-real eigenvalue. Since the sum and the product of the eigenvalues are real, it follows that the conjugate of the previous number is also an eigenvalue.\r\n\r\nHence, $P(A) = 0$, right?", + "Solution_4": "I don`t quite understand the last part of your solution... here`s mine:\r\nLet $\\alpha_1,\\alpha_2$ be the eigenvalues of $A$. Then the eigenvalues of $P(A)$ are $P(\\alpha_1)$ and $P(\\alpha_2)$.\r\nSo $\\det(P(A))=P(\\alpha_1)*P(\\alpha_2)=0$ wich implies $\\alpha_1,\\alpha_2$ are roots of $P$ and they are conjugated. But $\\alpha_1,\\alpha_2$ are roots of $P_A$ so $P_A/P$ equivalent to $P(X)=P_A(X)Q(X)$ and for $X=A$ we get $P(A)=O_2$.", + "Solution_5": "[quote=\"spix\"]Then the eigenvalues of $P(A)$ are $P(\\alpha_1)$ and $P(\\alpha_2)$.[/quote]\r\n\r\nI'm not very experienced with eigenvalues. Can $P(A)$ have other eigenvalues except those two??? (it's very easy to prove that those two are indeed eigenvalues of $P(A)$)\r\n\r\n\r\nWhat don't you understand in my solution? Maybe I did some stupid mistake... please tell me", + "Solution_6": "$P(A)$ is a $2x2$ matrix so it has two eigenvalues. There is nothing wrong it`s the same ideea... I did`nt understand how you jumped to that conclusion in the end... but now it`s clear to me... you just said a little more straightforward than me. Basically the solutions are the same... :)", + "Solution_7": "[quote=\"spix\"]\n\n Can this problem be generalised?[/quote]\r\n\r\nTheorem:\r\nK commutative field, $A\\in M_q(K)$ a matrix \r\n$P_1,...,P_q$ monic polynomials in K[X] such that they have no commun roots in $\\overline{K}$\r\nalgebraic closure of K\r\nwith $det(P_1(A))=....=det(P_q(A))=0$ \r\n\r\nThen $P_1(A)P_2(A)...P_q(A)=0$\r\n----------------------------------------------------------------\r\nProof:\r\nLet's work in $\\overline{K[X]}$ \r\n$P_1(X)=\\prod(X-a_k^{(1)})$\r\n...........................................\r\n$P_q(X)=\\prod(X-a_k^{(q)})$\r\n\r\n\r\n$0=det(P_1(A))=\\prod_{k}det(A-a_k^{(1)}I_n)$ \r\n..................................................................\r\n$0=det(P_1(A))=\\prod_{k}det(A-a_k^{(1)}I_n)$\r\n\r\nwe deduce there exist $a_{i_1}^{(1)},....,a_{i_q}^{(q)}$ distincts eigenvalues of A\r\n\r\n\r\n\r\n$(A-a_{i_1}^{(1)}I_n)...(A-a_{i_q}^{(q)}I_n)=0$ \r\n\r\n\r\nWe factor in $\\overline{K[X]}$\r\n$P_1(A)P_2(A)...P_q(A)=(A-a_{i_1}^{(1)}I_n)...(A-a_{i_q}^{(q)}I_n)W(A)=0$" +} +{ + "Tag": [ + "number theory unsolved", + "number theory" + ], + "Problem": "prove that the equation x^2-y^3=7 havn't any solutions such that (x,y) de N\u00b2", + "Solution_1": "Write it as $x^{2}+1=y^{3}+8=(y+2)(y^{2}-2y+4)$ and consider primes dividing $x^{2}+1$.\r\nBut the title \"eazy\" is really bad!", + "Solution_2": "whe the title is bad?", + "Solution_3": "Probably because the problem isnt that easy", + "Solution_4": "Probably because it's a useless and subjective information telling [b]nothing[/b] about the problem. See http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 .", + "Solution_5": "i m so soory" +} +{ + "Tag": [ + "analytic geometry", + "puzzles" + ], + "Problem": "I've being staring at this puzzle on and off for a week now and I'm getting nowhere. I'm usually pretty good at solving math related puzzles, but this one has me drawing a blank. \r\n\r\nhttp://www.geocaching.com/seek/cache_details.aspx?guid=fbede556-3bc1-4473-8109-a6c60d50a488\r\n\r\nI'm not looking for a outright solution as I do want to take some satisfaction in solving it, but if someone could hand hold me through some ideas that would be greatly appreciated.", + "Solution_1": "Ooh, that's a good one. Hmm........well, if you don't want an outright solution, perhaps you could draw it out?\r\nHope that helped.... :)", + "Solution_2": "I'm a little confused. What is the \"posted coordinate\", and what is the \"cache\"?" +} +{ + "Tag": [ + "AMC", + "AIME", + "articles" + ], + "Problem": "Blah see poll.", + "Solution_1": "hmm... anyone that has any political importance pronounces it Missouruh...", + "Solution_2": "Non-Missourians seem to pronounce Missouri \"Missouruh\" more than Missourians...\r\n\r\nThe only other prounciation I've heard of Missouri was \"misery,\" but that was by a person who moved from Maryland to Missouri...", + "Solution_3": "I say Missouri.", + "Solution_4": "How does ri translate to ruh", + "Solution_5": "Ask the person who chose that option.", + "Solution_6": "I've never heard anyone say it missouruh here.", + "Solution_7": "[quote]I've never heard anyone say it missouruh here.[/quote]\r\n\r\n\r\nMaybe because they never talk about Missouri in Conneticut? :D", + "Solution_8": "I always thought it was spelled Connecticut...\r\n\r\nWhy would Connecticuttians be discussing Missouri, and even if they were, how would they pronounce it?", + "Solution_9": "THis is getting confusing O_O", + "Solution_10": "Meh, sour, rye. (pronounce words)\r\n\r\nIs the proper pronounciationism.", + "Solution_11": "No, it's pronounced\r\n\r\nmuh-zir(rhymes with sir)-ee", + "Solution_12": "[quote]dragonlaird7 I always thought it was spelled Connecticut... \n[quote]\n\nAh, but technically Connecticut is typically pronounced Conn-et-i-cut, so I put use to that.\n\nI would not know how people in [b]Connecticut[/b] would pronounce Missouri, but I do not intend to fly a plane to Hartford and ask each civilian. \n\n[i]Ask ernie ?[/i][/quote][/quote]", + "Solution_13": "Accidentally clicked on quotes key just after sending message. You can figure it out...the first message is quote...my reply is the quoted box. :D :D :D", + "Solution_14": "Mees-your-hee.", + "Solution_15": "Mousuri is pronounced \"misery\"...its true!!!!(idk how to spell, and my spell check wont make missouri missouri)", + "Solution_16": "Ok, that was the oddest spelling of missouri ive ever seen.\r\n\r\nBy the way, AIME, what are you doing on the Missouri forum?", + "Solution_17": "How are you NOT in the Missouri forum?", + "Solution_18": "[quote=\"justinahmann\"][i]Ask ernie ?[/i][/quote]\r\n\r\nWe Connecticutians pronounce it \"cun-et-ih-cut\". :P\r\n\r\nAnd we also pronounce Missouri with its synonym, [i]misery[/i].", + "Solution_19": "[quote=\"pianoforte\"]Meh, sour, rye. (pronounce words)\n\nIs the proper pronounciationism.[/quote]\r\nMan over here in Cali we call it misery. Pun fully ****ing intended.", + "Solution_20": "CO-NN-ECTI-CU-T\r\n\r\nM-IS-SOUR-I\r\n\r\nNow pronounce it!", + "Solution_21": "Is M-IS-SOUR-I supposed to mean:\r\nem is sour aye", + "Solution_22": "That's how I say it...", + "Solution_23": "that actually sounds cool, then i thought", + "Solution_24": "u guys r confusing me. im just going to call it misery", + "Solution_25": "but that's wrong!", + "Solution_26": "Missouri is pr0nounced \"kuso\".", + "Solution_27": "evidence for \"kuso\"?", + "Solution_28": "\"Missouri is kuso.\"\r\n\r\n- Henry Clay", + "Solution_29": "[citation needed]", + "Solution_30": "How is it missouruh?", + "Solution_31": "Some radio broadcaster just called it \"Missourah\".", + "Solution_32": "its mizz-uhr-ee\r\n\r\nits like misery, but with emphasis on the zz soudn rather than the mmm sound. i think\r\n\r\nmizz-er-ee actually.", + "Solution_33": "why are you peeps even on this thread...?\r\n\r\nNo one from Missouri is posting on this thread, so none of you wouldn't know.", + "Solution_34": "well i was listening to the radio and my mom switched it to some economy thing and they were talking about how bad the recession was in \"missourah\"", + "Solution_35": "Interesting article:\r\n\r\n[url]http://missourifolkloresociety.truman.edu/Missouri%20Folklore%20Studies/THE%20PRONONCIATION%20OF%20MISSOURI.htm[/url]", + "Solution_36": "Mi (right between mee and meh)\r\nzur (as in azure)\r\nee\r\n\r\n-- New Yorker", + "Solution_37": "i thought it was \"tek-sas\"", + "Solution_38": "I say it Missouri because I'm from Missouri. I'm the only one in my class (Algebra I) on Mondays!!! Lol!", + "Solution_39": "Well, my social studies teacher says Missouruh and she's from Tennessee. She's probably the only one in Ohio that says it that way, though.", + "Solution_40": "It is pronounced\nmizz or E(like saying the letter E)", + "Solution_41": "It should be noted that despite the posts I have made in this thread, I do not have one of those dots saying I have done so...\n\nSOMETHING IS ROTTEN IN THE STATE OF MISSOURI!", + "Solution_42": "btw ILU ZOMG MEWTO YOU ARE SO COOL UNLOCK THE NO SPAM THREAD KK?" +} +{ + "Tag": [ + "inequalities", + "national olympiad" + ], + "Problem": "Where can I find past year national MO papers from around the world?\r\n\r\nFor example, China Mathematical Olympiad 2002. I wish to have a look at the 'superhard' inequality problem.\r\n\r\nThank you.", + "Solution_1": "Your user name shows that you are a Chinese Malaysian, so I guess that you might know Chinese.\r\nLook at this site: http://www.mathfan.net/Soft/jstd/gzjs/Index.html You can download the China TSTs from 1994 to 2003(no 2001 and 2000) with full solutions, you can also download some other wonderful articles.\r\nLook at this site: http://www.mathfan.net/Soft/jstd/imo/200405/16.html You can download the CMO problems from 1986 to 2004(1st to 19th) but no solution.\r\nMAYBE before you download SOME articles, you should log in first.\r\nEnjoy it!", + "Solution_2": "is there an option to see these sites in english?", + "Solution_3": "You can also check out [url]http://www.kalva.demon.co.uk/[/url] for problems from national olympiads from around the world.", + "Solution_4": "Or you can be lazy and just browser through the thousands of problems on this very site :P :)", + "Solution_5": "By the way, may I post some past olympiads that haven't posted here yet? For example, I have the problems of District Stage of All-Russian Olympiad for the several last years. Should I post it?", + "Solution_6": "I don't think anybody could possibly object to that. :)", + "Solution_7": "[quote=\"Remike\"]By the way, may I post some past olympiads that haven't posted here yet? For example, I have the problems of District Stage of All-Russian Olympiad for the several last years. Should I post it?[/quote]\r\n\r\nYeah, we would really appreicate it. Please start with the oldest problems. :)" +} +{ + "Tag": [ + "floor function", + "calculus", + "calculus computations" + ], + "Problem": "show :D \r\n\r\n$ \\sum_{k\\equal{}0}^{n}\\;\\frac{n!}{k!}\\;\\equal{}\\;\\boxed{\\lfloor\\;n!\\cdot e\\;\\rfloor}$", + "Solution_1": "(I'm assuming that $ n > 1$, since the problem is false for $ n \\equal{} 0$, and trivial for $ n \\equal{} 1$)\r\n\r\nThe series $ \\sum_{k\\equal{}n\\plus{}1}^{\\infty}\\frac{n!}{k!}$ converges by comparison to the geometric series. Furthermore, it converges to some number $ < 1$ since:\r\n\r\n$ \\sum_{k\\equal{}n\\plus{}1}^{\\infty}\\frac{n!}{k!}\\;\\leq\\;\\sum_{k\\equal{}1}^{\\infty}\\frac{1}{(n\\plus{}1)^{k}}\\; \\equal{}\\frac{1}{n}< 1$.\r\n\r\nFinally, noticing that $ \\lfloor n!\\cdot e\\rfloor \\equal{}\\sum_{k\\equal{}0}^{n}\\frac{n!}{k!}\\;\\plus{}\\left\\lfloor\\sum_{k\\equal{}n\\plus{}1}^{\\infty}\\frac{n!}{k!}\\right\\rfloor$ from the power series of $ \\exp x$, the result follows." +} +{ + "Tag": [ + "USAMTS" + ], + "Problem": "Anyone do 3D art? I use 3Dsmax. \r\n\r\n[img]http://img503.imageshack.us/img503/8730/viewport1yy.png[/img]\r\n\r\nMissing wing webbing, but thats not important :roll:", + "Solution_1": "Holy crud that is absolutley awesome! (make it look like Saphira or Thorn)(Eragon/eldest characters) the only kind of c.g. do are simple ms paint crap for school diagrams and simple USAMTS diagrams(curse you round 2!!)", + "Solution_2": "I've learned a little POV-Ray, but I never practice with it, so I've never done anything interesting. My most complex creation is somewhere on this page: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35620&postorder=asc&start=80 . (The final version is in my second post on the page.)", + "Solution_3": "Nice to meet you A+Math,I use $\\mathrm{3D\\ S\\ MAX}$,too.\r\n\r\nI can't see you picture :( ;)", + "Solution_4": "Hmm that looks familiar... might have seen it at CGtalk...", + "Solution_5": "I do post once in a while on CGTalk. I don't recall doing a WIP of the dragon however.", + "Solution_6": "nice, i use blender", + "Solution_7": "Anybody use SCED scene editor?\r\n\r\nI'm trying to download it, but I don't know how to \"gunzip\" or \"untar\" a winrar file. Anybody wanna help?\r\n\r\nBy the way, this is an interesting website:\r\n\r\n[url=http://images.google.com/imgres?imgurl=http://www.nada.kth.se/~asa/bilder/pattern.gif&imgrefurl=http://www.nada.kth.se/~asa/ray.html&h=384&w=512&sz=74&hl=en&start=21&tbnid=yY1HNuYeVR4-uM:&tbnh=98&tbnw=131&prev=/images%3Fq%3Dpattern%26start%3D20%26ndsp%3D20%26svnum%3D10%26hl%3Den%26lr%3D%26sa%3DN]I think this is a pretty cool place to see a demonstration of fine PovRay & Bryce 4 stuff[/url]\r\n\r\nHow did you ever get 3D S MAX? It costs so much..." +} +{ + "Tag": [], + "Problem": "1/a+1/(a+1)+1/(a+2)...+1/(ab+c) (you keep adding 1 to k until you reach ab+c)where a,b, and c are some integer. is it true that the above equation will get smaller if a increases but b and c stay the same?", + "Solution_1": "Clearly our \"worst case\" is when c = 0. Therefore, we wish to show that\r\n\r\n[tex]\\frac{1}{a} > \\frac{1}{ab+1}+\\frac{1}{ab+2}+\\frac{1}{ab+3}+ ... \\frac{1}{ab+b}[/tex]\r\n\r\nWe have\r\n\r\n[tex]\\frac{1}{a} = \\frac{b}{ab} > \\frac{1}{ab+1}+\\frac{1}{ab+2}+\\frac{1}{ab+3}+ ... \\frac{1}{ab+b}[/tex]\r\n\r\n[tex]QED[/tex]", + "Solution_2": "is there an equation for this?", + "Solution_3": "[quote=\"Scrambled\"]is there an equation for this?[/quote]\r\n\r\nErr... what is the \"this\"?", + "Solution_4": "[quote=\"probability1.01\"][quote=\"Scrambled\"]is there an equation for this?[/quote]\n\nErr... what is the \"this\"?[/quote]\r\n\r\nthis is the value of the equation. What i mean is if there is a formula for calculating how much the total value drops if a increases", + "Solution_5": "I don't think there is, but I could be missing something. Substitute ab = X\r\n\r\n[tex]\\mbox{Subtracting gives}[/tex]\r\n\r\n[tex]\\frac{1}{X+1}+\\frac{2}{X+2}+\\frac{3}{X+3}+ ... \\frac{b}{X+b}[/tex]\r\n\r\nThat's all I can get, not that it helps..." +} +{ + "Tag": [ + "ratio", + "geometry open", + "geometry" + ], + "Problem": "To prove without use cross ratio:\r\n\r\n[color=blue]Theorem. A triangle and its polar triangle are in perspective.[/color]\r\n\r\nthanks a lot\r\n\r\nJos\u00e9 Carlos", + "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=38235 (if you mean the same \"polar triangle\").\r\n\r\n darij" +} +{ + "Tag": [ + "geometry", + "geometry solved" + ], + "Problem": "Let $ABC$ be a triangle and $P$ point in its interior. Denote by $D,E,F$ feet of perpendicular from $P$ to the lines $BC,CA$ and $AB$ respectively. Suppose that $AP^2+PD^2=BP^2+PE^2=CP^2+PF^2$.Prove that $P$ is a center of the circle circumscribed about $\\triangle S_aS_bS_c$ , where $S_a,S_b,S_c$ are excenters of the triangle $ABC$.", + "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?p=18550 .\r\n\r\n darij" +} +{ + "Tag": [ + "geometry", + "incenter", + "circumcircle", + "geometry proposed" + ], + "Problem": "Let N,I the nine-point center and the incenter of a triangle.\r\nLet R,r denote the radii of its cimcumcircle and incircle.\r\nProve that \r\n\r\n NI=R/2 -r\r\n\r\nPlease try to prove it without using inversion.\r\n\r\nNow using Feuerbach Theorem you can prove this elegant fact.\r\n\r\nConsider the following 4 points of a triangle: the circumcenter,the incenter, the orthocenter and the nine-poit circle. Show that no three\r\nof these points can be the vertices of a nondegenerate equilateral triangle.", + "Solution_1": "It is well-known that the nine-point center N is the midpoint of [OH], where O is the circumcenter and H the orthocenter. It follows that the line (IN) is a median in the triangle OIH.\r\nIt follows that NI 2 = (OI 2 + HI 2 )/2 - OH 2 /4.\r\n\r\nNow, it suffices to use that :\r\nOI 2 = R 2 - 2Rr.\r\nHI 2 = 4R 2 + 4Rr + 3 r 2 - p 2 \r\nOH 2 = 9R - (a 2 +b 2 +c 2 ) = 9R 2 - (2p 2 - 2r 2 - 8Rr),\r\nwhere p denotes the semi-perimeter.\r\n\r\nPierre.", + "Solution_2": "You can prove NI=R/2-r [b]using[/b] Feuerbach. The 9point circle and the incircle touch at a point Q lying on NI. We have NQ=R/2 and IQ=r, so we get the desired result.\r\n\r\nSuppose one of the triangles is equilateral.\r\nIf R=2r then OI=0, so O=I so the triangle in equilateral, so O,I,N,H coincide.\r\nIf R>2r, then OI>2*NI (just use the formulas).\r\nSo NOI is not equilateral. HNO is not equilateral because the 3 points are collinear. SO HIN must be eqiolateral.\r\nBecause N is the midpoint of OH, we must have OI/ \\sqrt3 =NI=HN=NO.\r\nSo OI=IN* \\sqrt 3. But OI>2*NI. So \\sqrt 3>2 (a contradiction)." +} +{ + "Tag": [ + "\\/closed" + ], + "Problem": "I have a few questions but dont know where to look...\r\n\r\n1. How do you delete topics or posts you make by mistake?\r\n\r\n2. How do you create an avatar?\r\n\r\nPlease don't laugh at me because i haven't used AoPS much... :(", + "Solution_1": "1. Regular users cannot delete topics. To delete a post, click the X in the bottom-right corner (disappears once somebody posts after you).\r\n\r\n2. http://www.artofproblemsolving.com/Forum/usercp.php?mode=editprofile&sub=avatar", + "Solution_2": "Is it possible to edit a GIF image so that it is no more than 80 pixels wide and 100 pixels high?", + "Solution_3": "Yes, and if you don't know how, I'll be glad to do it for you.", + "Solution_4": "Never mind, I found an avatar that fits it.\r\n\r\n<-----See? Its a Sonic!\r\n\r\nThanks for your help.", + "Solution_5": "[quote=\"i_like_pie\"]1. Regular users cannot delete topics. To delete a post, click the X in the bottom-right corner (disappears once somebody posts after you)[/quote]\r\n\r\nTo delete one of your own topics, delete your own first post in the topic. This may not be possible, though, depending on the forum it was posted in.", + "Solution_6": "Actually, it is best to contact a moderator if you would like anything deleted; they can decide if it should be deleted or not and can always do so." +} +{ + "Tag": [ + "MATHCOUNTS", + "FTW" + ], + "Problem": "My state competition is this Saturday any last minute cramming suggestions?", + "Solution_1": "just do problems/tests (old nats seem to work well as i've heard from others). make sure to relax the night before. being nervous killed me at state this year (5th... :( ) i made about 10 careless mistakes. any one of them made correct would have brought me up to 4th or higher.\r\n\r\nEDIT: Think of it this way. You can only do as well as you--well, can do. Just try your best--you can't do better than that. That's all that anyone expects of you. A lucky charm might help too... :D \r\n\r\nAnyway, good luck!", + "Solution_2": "Do all of the previous mathcounts state competitions (2001-2007) and most importantly RELAX. Don't get nervous, it will bring your score done.\r\n\r\nEDIT: basically samething as math154", + "Solution_3": "do you live in pennsylvannia. i would like to know.\r\n\r\nbtw, how many prims are less than 25. i would like to know.", + "Solution_4": "i \"don't know\" how many primes are less than 25. If you want a prime number, I think $ \\LARGE\\phi$ is in its prime.\r\n\r\n\"Show Me!!!\" :D", + "Solution_5": "No I live in mississippi.... But my State's date is the same as yours... and theres 9 primes less than 25 (right?)", + "Solution_6": "yeah i think there are. My calculator's \"Prime Generator\" says so.", + "Solution_7": "my bad, i meant 100. i THINK the answer is 25...right?", + "Solution_8": "[quote=\"abacadaea\"]my bad, i meant 100. i THINK the answer is 25...right?[/quote]\nyeah my p-gen says so, and i memorized that, too.\n\n[quote=\"NightNinja94\"]Btw math154 Ive got a lucky Jacket...[/quote]\r\nThat's good. Just make sure to not keep it on if it's burning hot.", + "Solution_9": "Don't cram.\r\nJust review a test a day, and make sure you get enough sleep on the night before.", + "Solution_10": "Review like the 2007 or 2006 State Test or something recent.", + "Solution_11": "Yeah on the night before review a recent test, and be sure to bring some practice materials with you to the contest so you feel ready when the real test comes.", + "Solution_12": "Don't bother learning any new concepts; it's too late for that. Your time is better spent reviewing trouble spots and working on old problems.", + "Solution_13": "I think it's too late to work on old problems, unless you have a day off tomorrow.\r\nI would just go do random problems on saab or do that arithmetic game thing.", + "Solution_14": "What \"arithmetic game thing\"?", + "Solution_15": "I think xp math is talking about FTW?", + "Solution_16": "no, it's different", + "Solution_17": "what then? math mayhem(record:1640 pwn3d!!)?", + "Solution_18": "math mayhem is totally awesome!!! especially multiplication. i hate when i typo... :(", + "Solution_19": "[quote=\"math154\"]What \"arithmetic game thing\"?[/quote]\r\n\r\na fun and popular math game until FTW came along and smashed it into pieces, and no one ever bother to pick up the pieces....", + "Solution_20": "I don't go on FTW much anymore. It's kinda a waste of time if you can mem a lot of the questions there. :wink:", + "Solution_21": "Thanks guys my states tommorow moorning thanks for all the help", + "Solution_22": "Well, again, good luck to everyone!", + "Solution_23": "old nuts are really good and also try some chocolate too. it always helps me :)\r\n\r\ngood luck !", + "Solution_24": "[quote=\"Anikaa\"]old nuts are really good and also try some chocolate too. it always helps me :)\n\ngood luck ![/quote]\r\n\r\nYes, I will assume chocolate is what made me get 3rd in states!\r\n...\r\n...\r\n...\r\nYou never know... it could've! :D", + "Solution_25": "What place were you expecting, yongyi?", + "Solution_26": "[quote=\"Yongyi781\"][quote=\"Anikaa\"]old nuts are really good and also try some chocolate too. it always helps me :)\n\ngood luck ![/quote]\n\nYes, I will assume chocolate is what made me get 3rd in states!\n...\n...\n...\nYou never know... it could've! :D[/quote]\r\nHow would eating chocolate really help you get a good score at states? It didn't work for me. It [i]hurt[/i] me.", + "Solution_27": "Yeah. Sugar gives you a burst...but then you crash. So basically, you'll fail the Target Round if you eat chocolate...I think...", + "Solution_28": "just eat pasta the night before. Keep a steady supply of chocolate to eat during the test and before it.\r\n\r\nHonestly, the way I do well is that I practice and try hard.", + "Solution_29": "Ihatepie, the Texas state competition is going on right now...nobody can use your great tips :D \r\nOooohhh..........who will make it? The anticipation...", + "Solution_30": "i guarantee that all of you know enough concepts for mc. Problem solving is something that cannot be learned, you only develop it.", + "Solution_31": "well we'll see who wins. I put 20 dollars on Bobby Shen for taking first prize. :D", + "Solution_32": "[quote=\"NightNinja94\"]Any last minute cramming suggestions?[/quote]\r\n\r\nNO NO NO DON'T CRAM! CRAM EARLY, RELAX LATER!", + "Solution_33": "Well what I do is Do my test regularly. Leave one full year for the night before. I do team, countdown, sprint, and target in that order, then go to sleep at around 9 (I usually at 11.) This doesn't work if you are one of the people who stay up all night thinking about the contest.\r\n\r\nRemember, don't get nervous. Nothing will happen if you lose.", + "Solution_34": "What you have to remember is not to get stressed out or nervous. MC isn't your life!!!", + "Solution_35": "For some, it is...", + "Solution_36": "Lots of people stay up really late a few days before the competition preparing, and then try to sleep early the night before. It doesn't work that way. They usually just stay awake in bed. The point I'm trying to make is that you should sleep early the last few days before the competition to actually be refreshed.", + "Solution_37": "I actually made that mistake this year too. I should've rested more. Because even though I was doing so many problems, it didn't seem to help. And actually, sleep is directly correlated to memory. So if you don't get enough sleep...you could forget 3*8=24.", + "Solution_38": "The reason why people fail something, is that thay are too nervous at that time ,I think. To take something easy, is the best way to do what you want. For example I'm always well prepared for my history classes , but very often I forget dates and things I had to say. That happens because I'm too nervous at that moment. \r\n\r\nAnd at last you must belive in what are you doing, and everything will be okay. \r\n\r\n P.S. chocolate helps me to stay calm and do not worry too much :) \r\n\r\nI do not think I explained clearly what I meant but...", + "Solution_39": "[quote=\"Ihatepie\"]well we'll see who wins. I put 20 dollars on Bobby Shen for taking first prize. :D[/quote]\r\nYeah, so do I. Just as Kevin Chen said, Bobby will win!", + "Solution_40": "I meant at the state competition. But he has a really good chance at nats. Who knows, he might win.", + "Solution_41": "always, and I mean ALWAYS eat hard-boiled eggs in between Target round questions. it helps.", + "Solution_42": "People have different preferences. I personally don't like eggs...and the main thing you have to do is do the easy question, check it, and do the hard one in target round. For sprint, you just have to do first 10 in 5 mins and then work through the rest like 1.5 mins per question." +} +{ + "Tag": [ + "AMC", + "AIME", + "search", + "function" + ], + "Problem": "Did most AoPSers take AIME I or AIME II?", + "Solution_1": "The search function usually helps a lot...\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28592]www.artofproblemsolving.com/Forum/viewtopic.php?t=28592[/url]", + "Solution_2": "Pardon me, Ms. Perfection. Should I delete this poll?", + "Solution_3": "If Ms. Perfection is aimed at that wonderful picture of BoA i have for an avatar, sure...I'm pretty sure she'd want you to delete this. *nods*\r\n\r\nOtherwise, it doesn't really matter to me...", + "Solution_4": "Apparently, I don't have the privileges to delete the poll (only a moderator or adminsitrator can do that now that votes have been cast). Sorry. But seeing that your a moderator, you probably can. \r\n\r\n(And no, Ms. Perfection was supposed to be directed at you because I thought your avatar was your picture. I don't know who BoA is). :)", + "Solution_5": "BoA stands for beat of angel, I think she is a Korean singer.", + "Solution_6": "Oh gj eryaman. GJ indeed :-) However, she sings in Korean, Japanese, English, and Chinese. At 18, she's basically the pop princess of Asia. Oh...and she's a model. And she dances well. And...yea...she does just about everything well. :-D", + "Solution_7": "[quote=\"bubala\"]Pardon me, Ms. Perfection. Should I delete this poll?[/quote]\r\n\r\nI find this entertaining... you called Miths a Ms... haha. :D", + "Solution_8": "hey hey you...lol...shuddap. meanie. lol I like my avatar...", + "Solution_9": "Miths=Miss :P :P :P", + "Solution_10": "i like your avatar too miths. boa is hot as hell.", + "Solution_11": "[quote=\"DPopov\"]Miths=Miss :P :P :P[/quote]\r\n\r\nYou do not want to be pwned irl and you know it. :P I know...we'll settle it with a game of bridge.\r\n\r\nAnd I do appreciate that gesture codeblue: finally, someone who embodies the ideals of masculine hormones.", + "Solution_12": "[quote=\"codeblue87\"]i like your avatar too miths. boa is hot as -.[/quote]\r\n\r\nYou won't find too much disagreement with that statement." +} +{ + "Tag": [ + "floor function", + "inequalities", + "algebra unsolved", + "algebra" + ], + "Problem": "Determine a simple continuous fraction of $\\frac{F_{10n+1}}{F_{10n}}$ where $F_{k}$ is k-th Fibonacci number.\r\n\r\nPs: what is a simple continuous fraction?", + "Solution_1": "A simple continuous fraction is of the form\r\n$a_{1}+\\frac{1}{a_{2}+\\frac{1}{a_{3}+\\frac{1}{...+\\frac{1}{a_{n}}}}}$\r\nwhere the $a_{i}$'s are positive integers. This simple continuous fraction is denoted $[a_{1},a_{2},...,a_{n}]$. It is easy to see that each rational number $k$ can be represented as a simple continuous fraction in exactly one unique way. \r\n\r\nTo get to this representation, we first notice $a_{1}=\\lfloor k \\rfloor$. Similarly, $a_{2}=\\lfloor \\frac{1}{k-a_{1}}\\rfloor$, and so on. Each step, we take the largest integer that fits into the number, and store it as $a_{i}$. We then subtract it from the number, and raise the number to the power of negative one. the process is repeated until $0$ is reached. \r\n\r\nNow consider $\\frac{F_{n+1}}{F_{n}}$. This is the same as $\\frac{F_{n}+F_{n-1}}{F_{n}}=1+\\frac{F_{n-1}}{F_{n}}$\r\nwhen the process described above is repeated, a sequence of $n+1$ ones is produced.\r\n\r\nIn the case of $10n+1$, we get the sequence $[1,1,1,...,1]$, with $10n+1$ ones.", + "Solution_2": "[quote=\"shalevbd\"]It is easy to see that each rational number $k$ can be represented as a simple continuous fraction in exactly one unique way.[/quote]\r\nYou forgot mention that for this the inequality $a_{n}\\le 2$ must hold.\r\nExamples $[1,2]=[1,1,1]$\r\n...\r\nand\r\n$[1,1,...,1]$ with $10n+1$ ones equals $[1,1,...,1,2]$ with $10n$ ones." +} +{ + "Tag": [ + "inequalities unsolved", + "inequalities" + ], + "Problem": "Let $x;y;z;t$ be positive real numbers satisfying:$x+y+z+t=2$\r\nFind min $A=\\frac{(x+y+z)(x+y)}{xyzt}$", + "Solution_1": "I think, the answer is 16.", + "Solution_2": "Can you post you solution?", + "Solution_3": "[quote=\"chien than\"]Can you post you solution?[/quote]\r\nIt's very ugly:\r\n$\\frac{(x+y+z)(x+y)}{xyzt}\\geq16\\Leftrightarrow(x+y+z)(x+y)(x+y+z+t)^{2}\\geq64xyzt\\Leftrightarrow$\r\n$\\Leftrightarrow(x+y+z)(x+y)t^{2}-(64xyz-2(x+y+z)^{2}(x+y))t+(x+y+z)^{3}(x+y)\\geq0.$\r\nHence, remain to prove that \r\n$\\left(32xyz-(x+y+z)^{2}(x+y)\\right)^{2}-(x+y+z)^{4}(x+y)^{2}\\leq0\\Leftrightarrow$\r\n$\\Leftrightarrow(x+y+z)^{2}(x+y)\\geq16xyz\\Leftrightarrow(x+y)z^{2}-(16xy-2(x+y)^{2})z+(x+y)^{3}\\geq0.$\r\nHence, remain to prove that $\\left(8xy-(x+y)^{2}\\right)^{2}-(x+y)^{4}\\leq0\\Leftrightarrow(x-y)^{2}\\geq0.$\r\nEquality when $x=y=\\frac{1}{4},$ $z=\\frac{1}{2}$ and $t=1.$" +} +{ + "Tag": [ + "modular arithmetic" + ], + "Problem": "Prove that: p is prime, $ p\\geq 3$, the equation $ x^2 \\plus{} 1\\equiv\\pmod p$ have solution if $ p \\equal{} 4k \\plus{} 1$", + "Solution_1": "[quote=\"mitdacvn\"]Prove that: p is prime, $ p\\geq 3$, the equation $ x^2 \\plus{} 1(\\mod p)$ have solution if $ p \\equal{} 4k \\plus{} 1$[/quote]\r\nDo you mean $ x^2 \\plus{} 1 \\equiv 0 \\pmod{p}$?\r\nIf so, the well-known solution is to [hide=\"consider\"]$ x \\equal{} \\left(\\frac {p \\minus{} 1}{2}\\right)!$, and see why it works with Wilson's theorem.[/hide]\nEDIT: Or [hide=\"consider\"]$ g^k$, where $ g$ is a primitive root mod p. It works because $ g^{2k}$ has multiplicative order 2 and thus must be equivalent to $ \\minus{}1 \\pmod{p}$.[/hide]", + "Solution_2": "[quote=\"Brut3Forc3\"][quote=\"mitdacvn\"]Prove that: p is prime, $ p\\geq 3$, the equation $ x^2 \\plus{} 1(\\mod p)$ have solution if $ p \\equal{} 4k \\plus{} 1$[/quote]\nDo you mean $ x^2 \\plus{} 1 \\equiv 0 \\pmod{p}$?\nIf so, the well-known solution is to [hide=\"consider\"]$ x \\equal{} \\left(\\frac {p \\minus{} 1}{2}\\right)!$, and see why it works with Wilson's theorem.[/hide]\nEDIT: Or [hide=\"consider\"]$ g^k$, where $ g$ is a primitive root mod p. It works because $ g^{2k}$ has multiplicative order 2 and thus must be equivalent to $ \\minus{} 1 \\pmod{p}$.[/hide][/quote]\r\n\r\nEDIT: \r\nProve that: p is prime, $ p\\geq 3$, the equation $ x^2 \\plus{} 1\\equiv\\pmod p$ have solution [b]if and only if[/b] $ p \\equal{} 4k \\plus{} 1$\r\n\r\nGoing left, I can prove\r\nGoing right, I trying...\r\nWho can help me?", + "Solution_3": "It's much easier. Assume $ p \\equal{} 4k \\plus{} 3$, then obviously $ p$ does not divide $ x$ so\r\n\\[ x^2\\equiv \\minus{} 1\\implies1\\equiv x^{p \\minus{} 1}\\equiv x^{2\\cdot\\frac {p \\minus{} 1}{2}} \\equal{} ( \\minus{} 1)^{\\frac {p \\minus{} 1}{2}} \\equal{} \\minus{} 1\\pmod p.\\]" +} +{ + "Tag": [], + "Problem": "Is there such a whole number that starts with the digit 7, and it triples when you move its first digit to the end of the number?", + "Solution_1": "[hide=\"Answer\"]No. The first digit must be $1$, $2$, or $3$ for this to be possible, because the number of digits stays the same, and to triple a number that begins with a $7$, you must add a digit.[/hide]", + "Solution_2": "[hide]If the first digit is 7, when it is tripled it adds an extra digit making it impossible[/hide]\r\n\r\n[size=75][color=darkblue]